2012 ACM/ICPC 亚洲区 金华站
2024-05-11 12:00:53
题目链接 2012金华区域赛
Problem A
按a/b从小到大的顺序排队进行体检即可
1 #include<iostream>
2 #include<cstdio>
3 #include<cmath>
4 #include<cstdlib>
5 #include<algorithm>
6 #include<cstring>
7 #include<string>
8 #include<vector>
9 #include<map>
10 #include<set>
11 #include<queue>
12 using namespace std;
13 typedef long long ll;
14 const ll mod=365*24*60*60;
15 struct ss
16 {
17 int x,y;
18 double p;
19 };
20 ss a[100010];
21 int n;
22 inline bool cmp(ss a,ss b)
23 {
24 return a.p<b.p;
25 }
26 int main()
27 {
28 while (~scanf("%d",&n))
29 {
30 if (n==0) break;
31 int i;
32 for (i=1;i<=n;i++)
33 {
34 scanf("%d%d",&a[i].x,&a[i].y);
35 a[i].p=a[i].x*1.0/a[i].y;
36 }
37 sort(a+1,a+n+1,cmp);
38 ll sum=0;
39 for (i=1;i<=n;i++)
40 sum=((sum+sum*a[i].y%mod)%mod+a[i].x)%mod;
41 printf("%lld\n",sum);
42 }
43 return 0;
44 }View Code
Problem B
Problem C
先把所有的坐标离散化
将起点、终点和所有矩形的点
每个点4个方向，拆成4个点
然后如果一个点的一个方向能经过一次转弯直接到另一个点
就连一条边
建完图后用SPFA求最短路即可
但本题细节较多
最难处理的是墙角的转角情况
可以先把所有墙角预处理出来
一共8种情况
其他的就离散化后坐标乘2
然后用个数组标记每个点会经过几次
如果点已经在矩形内部
就直接置为经过2次
然后在判断时 判断路线上有没有经过2次以上的点
或者恰有一个墙角点但可以转弯
1 #include<iostream>
2 #include<cstdio>
3 #include<cmath>
4 #include<cstdlib>
5 #include<algorithm>
6 #include<cstring>
7 #include<string>
8 #include<vector>
9 #include<map>
10 #include<set>
11 #include<queue>
12 using namespace std;
13 #define y1 hjhk
14 #define y2 mhgu
15 const int inf=(1<<31)-1;
16 struct Hash : vector<int> {
17 void prepare() {
18 sort(begin(), end());
19 erase(unique(begin(), end()), end());
20 }
21 int get(int x) {
22 return lower_bound(begin(), end(), x)-begin()+1;
23 }
24 } has;
25 int x1,y1,x2,y2,n;
26 struct ss
27 {
28 int x1,y1,x2,y2;
29 };
30 ss a[51];
31 int dis[1500][4];
32 int can[1500][1500];
33 bool mop[1500][1500][4][4];
34 queue<pair<int,int>> q;
35 bool vis[1500][4];
36 pair<int,int> det[1500];
37 void relax(pair<int,int> from,int x2,int y2,int ndis)
38 {
39 //cout<<from.first<<" "<<from.second<<" "<<x2<<" "<<y2<<" "<<ndis<<endl;
40 if (ndis<dis[x2][y2])
41 {
42 dis[x2][y2]=ndis;
43 //cout<<from.first<<" "<<from.second<<" "<<x2<<" "<<y2<<" "<<ndis<<endl;
44 if (!vis[x2][y2])
45 {
46 q.push({x2,y2});
47 vis[x2][y2]=true;
48 }
49 }
50 }
51 int main()
52 {
53 //freopen("4444.in","r",stdin);
54 //freopen("out.txt","w",stdout);
55 while (~scanf("%d%d%d%d",&x1,&y1,&x2,&y2))
56 {
57 if (x1==0&&y1==0&&x2==0&&y2==0) return 0;
58 scanf("%d",&n);
59 int i,j;
60 for (i=1;i<=n;i++)
61 {
62 scanf("%d%d%d%d",&a[i].x1,&a[i].y1,&a[i].x2,&a[i].y2);
63 if (a[i].x1>a[i].x2)
64 swap(a[i].x1,a[i].x2);
65 if (a[i].y1>a[i].y2)
66 swap(a[i].y1,a[i].y2);
67 }
68 has.clear();
69 has.push_back(x1);
70 has.push_back(x2);
71 for (i=1;i<=n;i++)
72 {
73 has.push_back(a[i].x1);
74 has.push_back(a[i].x2);
75 }
76 has.prepare();
77 x1=has.get(x1)*2;
78 x2=has.get(x2)*2;
79 for (i=1;i<=n;i++)
80 {
81 a[i].x1=has.get(a[i].x1)*2;
82 a[i].x2=has.get(a[i].x2)*2;
83 }
84 has.clear();
85 has.push_back(y1);
86 has.push_back(y2);
87 for (i=1;i<=n;i++)
88 {
89 has.push_back(a[i].y1);
90 has.push_back(a[i].y2);
91 }
92 has.prepare();
93 y1=has.get(y1)*2;
94 y2=has.get(y2)*2;
95 for (i=1;i<=n;i++)
96 {
97 a[i].y1=has.get(a[i].y1)*2;
98 a[i].y2=has.get(a[i].y2)*2;
99 }
100 //cout<<x1<<" "<<y1<<" "<<x2<<" "<<y2<<endl;
101 //for (i=1;i<=n;i++)
102 // printf("%d %d %d %d\n",a[i].x1,a[i].y1,a[i].x2,a[i].y2);
103 memset(mop,false,sizeof(mop));
104 for (i=1;i<=n;i++)
105 for (j=1;j<=n;j++)
106 if (i!=j)
107 {
108 if (a[i].x2==a[j].x1&&a[i].y1<=a[j].y1)
109 {
110 mop[a[j].x1][a[j].y1][1][2]=true;
111 mop[a[j].x1][a[j].y1][0][3]=true;
112 //cout<<a[j].x1<<" "<<a[j].y1<<" "<<1<<" "<<2<<endl;
113
114 }
115 if (a[i].y2==a[j].y1&&a[i].x2<=a[j].x2)
116 {
117 mop[a[i].x2][a[i].y2][1][2]=true;
118 mop[a[i].x2][a[i].y2][0][3]=true;
119 //cout<<a[i].x2<<" "<<a[i].y2<<" "<<1<<" "<<2<<endl;
120 }
121 if (a[i].x2==a[j].x1&&a[i].y2<=a[j].y2)
122 {
123 mop[a[i].x2][a[i].y2][2][1]=true;
124 mop[a[i].x2][a[i].y2][3][0]=true;
125 //cout<<a[i].x2<<" "<<a[i].y2<<" "<<2<<" "<<1<<endl;
126 }
127 if (a[i].y2==a[j].y1&&a[i].x1<=a[j].x1)
128 {
129 mop[a[j].x1][a[j].y1][2][1]=true;
130 mop[a[j].x1][a[j].y1][3][0]=true;
131 //cout<<a[j].x1<<" "<<a[j].y1<<" "<<2<<" "<<1<<endl;
132 }
133 if (a[i].x2==a[j].x1&&a[i].y1>=a[j].y1)
134 {
135 mop[a[i].x2][a[i].y1][1][0]=true;
136 mop[a[i].x2][a[i].y1][2][3]=true;
137 //cout<<a[i].x2<<" "<<a[i].y1<<" "<<1<<" "<<0<<endl;
138 }
139 if (a[i].y2==a[j].y1&&a[i].x1>=a[j].x1)
140 {
141 mop[a[i].x1][a[i].y2][1][0]=true;
142 mop[a[i].x1][a[i].y2][2][3]=true;
143 //cout<<a[i].x1<<" "<<a[i].y2<<" "<<1<<" "<<0<<endl;
144 }
145 if (a[i].y2==a[j].y1&&a[i].x2>=a[j].x2)
146 {
147 mop[a[j].x2][a[j].y1][0][1]=true;
148 mop[a[j].x2][a[j].y1][3][2]=true;
149 //cout<<a[j].x2<<" "<<a[j].y1<<" "<<0<<" "<<1<<endl;
150 }
151 if (a[i].x2==a[j].x2&&a[i].y2>=a[j].y2)
152 {
153 mop[a[j].x1][a[j].y2][0][1]=true;
154 mop[a[j].x1][a[j].y2][3][2]=true;
155 //cout<<a[j].x1<<" "<<a[j].y2<<" "<<0<<" "<<1<<endl;
156 }
157 }
158 memset(can,0,sizeof(can));
159 //cout<<x1<<" "<<y1<<" "<<x2<<" "<<y2<<endl;
160 for (i=1;i<=n;i++)
161 {
162 for (int x=a[i].x1;x<=a[i].x2;x++)
163 for (int y=a[i].y1;y<=a[i].y2;y++)
164 {
165 can[x][y]++;
166 if (x!=a[i].x1&&x!=a[i].x2&&y!=a[i].y1&&y!=a[i].y2)
167 {
168 can[x][y]=2;
169 //if (x==11&&y==60)
170 // cout<<a[i].x1<<" "<<a[i].x2<<" "<<a[i].y1<<" "<<a[i].y2<<endl;
171 }
172 if (can[x][y]>2) can[x][y]=2;
173 //cout<<x<<" "<<y<<" "<<can[x][y]<<endl;
174 }
175 }
176 for (i=1;i<=n;i++)
177 for (j=1;j<=n;j++)
178 if (i!=j)
179 {
180 if (a[i].x2==a[j].x1&&a[i].y1==a[j].y1) can[a[i].x2][a[i].y1]=1;
181 if (a[i].x2==a[j].x1&&a[i].y2==a[j].y2) can[a[i].x2][a[i].y2]=1;
182 if (a[i].y2==a[j].y1&&a[i].x1==a[j].x1) can[a[i].x1][a[i].y2]=1;
183 if (a[i].y2==a[j].y1&&a[i].x2==a[j].x2) can[a[i].x2][a[i].y2]=1;
184 }
185 int cnt=0;
186 det[0].first=x1;
187 det[0].second=y1;
188 for (i=1;i<=n;i++)
189 {
190 det[++cnt].first=a[i].x1;
191 det[cnt].second=a[i].y1;
192 det[++cnt].first=a[i].x1;
193 det[cnt].second=a[i].y2;
194 det[++cnt].first=a[i].x2;
195 det[cnt].second=a[i].y1;
196 det[++cnt].first=a[i].x2;
197 det[cnt].second=a[i].y2;
198 }
199 det[++cnt].first=x2;
200 det[cnt].second=y2;
201 for (i=0;i<=cnt;i++)
202 dis[i][0]=dis[i][1]=dis[i][2]=dis[i][3]=inf;
203 dis[0][0]=0;
204 dis[0][1]=0;
205 dis[0][2]=0;
206 dis[0][3]=0;
207 memset(vis,false,sizeof(vis));
208 vis[0][0]=true;
209 vis[0][1]=true;
210 vis[0][2]=true;
211 vis[0][3]=true;
212 while (!q.empty()) q.pop();
213 q.push({0,0});
214 q.push({0,1});
215 q.push({0,2});
216 q.push({0,3});
217 while (!q.empty())
218 {
219 pair<int,int> x=q.front();
220 q.pop();
221 vis[x.first][x.second]=false;
222 //cout<<det[x.first].first<<" "<<det[x.first].second<<endl;
223 //cout<<x.second<<endl;
224 //cout<< dis[x.first][x.second]<<endl;
225 for (i=0;i<=cnt;i++)
226 if (i!=x.first)
227 {
228 int x1=det[x.first].first;
229 int y1=det[x.first].second;
230 int x2=det[i].first;
231 int y2=det[i].second;
232 //cout<<i<<" "<<x1<<" "<<y1<<" "<<x2<<" "<<y2<<endl;
233 if (x.second==3)
234 {
235 if (x2==x1&&y2<y1)
236 {
237 bool fg=true;
238 for (int y=y2;y<=y1;y++)
239 if (can[x1][y]==2)
240 {
241 fg=false;
242 break;
243 }
244 if (fg)
245 relax(x,i,3,dis[x.first][x.second]);
246 }
247 else if (x2<x1&&y2<y1)
248 {
249 int fg=0;
250 for (int y=y2;y<=y1;y++)
251 if (can[x1][y]==2)
252 {
253 fg++;
254 }
255 for (int x=x2;x<=x1;x++)
256 if (can[x][y2]==2)
257 {
258 fg++;
259 }
260 if (fg==0||(fg==2&&can[x1][y2]==2&&mop[x1][y2][3][0]))
261 relax(x,i,0,dis[x.first][x.second]+1);
262 }
263 else if (x2>x1&&y2<y1)
264 {
265 int fg=0;
266 for (int y=y2;y<=y1;y++)
267 if (can[x1][y]==2)
268 {
269 fg++;
270 //break;
271 }
272 for (int x=x1;x<=x2;x++)
273 if (can[x][y2]==2)
274 {
275 fg++;
276 //break;
277 }
278 if (fg==0||(fg==2&&can[x1][y2]==2&&mop[x1][y2][3][2]))
279 relax(x,i,2,dis[x.first][x.second]+1);
280 }
281 relax(x,x.first,2,dis[x.first][x.second]+1);
282 relax(x,x.first,0,dis[x.first][x.second]+1);
283 }
284 else if (x.second==0)
285 {
286 if (x2<x1&&y2==y1)
287 {
288 bool fg=true;
289 for (int x=x2;x<=x1;x++)
290 if (can[x][y2]==2)
291 {
292 fg=false;
293 break;
294 }
295 if (fg)
296 relax(x,i,0,dis[x.first][x.second]);
297 }
298 else if (x2<x1&&y2<y1)
299 {
300 int fg=0;
301 for (int y=y2;y<=y1;y++)
302 if (can[x2][y]==2)
303 {
304 fg++;
305 //break;
306 }
307 for (int x=x2;x<=x1;x++)
308 if (can[x][y1]==2)
309 {
310 fg++;
311 //break;
312 }
313 if (fg==0||(fg==2&&can[x2][y1]==2&&mop[x2][y1][0][3]))
314 relax(x,i,3,dis[x.first][x.second]+1);
315 }
316 else if (x2<x1&&y2>y1)
317 {
318 int fg=0;
319 for (int y=y1;y<=y2;y++)
320 if (can[x2][y]==2)
321 {
322 fg++;
323 //break;
324 }
325 for (int x=x2;x<=x1;x++)
326 if (can[x][y1]==2)
327 {
328 fg++;
329 //break;
330 }
331 if (fg==0||(fg==2&&can[x2][y1]==2&&mop[x2][y1][0][1]))
332 relax(x,i,1,dis[x.first][x.second]+1);
333 }
334 relax(x,x.first,1,dis[x.first][x.second]+1);
335 relax(x,x.first,3,dis[x.first][x.second]+1);
336
337 }
338 else if (x.second==1)
339 {
340 if (x2==x1&&y2>y1)
341 {
342 bool fg=true;
343 for (int y=y1;y<=y2;y++)
344 if (can[x2][y]==2)
345 {
346 fg=false;
347 break;
348 }
349 if (fg)
350 relax(x,i,1,dis[x.first][x.second]);
351 }
352 else if (x2<x1&&y2>y1)
353 {
354 int fg=0;
355 for (int y=y1;y<=y2;y++)
356 if (can[x1][y]==2)
357 {
358 fg++;
359 //break;
360 }
361 for (int x=x2;x<=x1;x++)
362 if (can[x][y2]==2)
363 {
364 fg++;
365 //break;
366 }
367 if (fg==0||(fg==2&&can[x1][y2]==2&&mop[x1][y2][1][0]))
368 relax(x,i,0,dis[x.first][x.second]+1);
369 }
370 else if (x2>x1&&y2>y1)
371 {
372 int fg=0;
373 for (int y=y1;y<=y2;y++)
374 if (can[x1][y]==2)
375 {
376 fg++;
377 //break;
378 }
379 for (int x=x1;x<=x2;x++)
380 if (can[x][y2]==2)
381 {
382 fg++;
383 //break;
384 }
385 if (fg==0||(fg==2&&can[x1][y2]==2&&mop[x1][y2][1][2]))
386 relax(x,i,2,dis[x.first][x.second]+1);
387 }
388 relax(x,x.first,2,dis[x.first][x.second]+1);
389 relax(x,x.first,0,dis[x.first][x.second]+1);
390
391 }
392 else if (x.second==2)
393 {
394 if (x2>x1&&y2==y1)
395 {
396 bool fg=true;
397 for (int x=x1;x<=x2;x++)
398 if (can[x][y2]==2)
399 {
400 //cout<<x<<" "<<y2<<endl;
401 fg=false;
402 break;
403 }
404 //cout<<fg<<" "<<cnt<<" "<<i<<endl;
405 if (fg)
406 relax(x,i,2,dis[x.first][x.second]);
407 }
408 else if (x2>x1&&y2<y1)
409 {
410 int fg=0;
411 for (int y=y2;y<=y1;y++)
412 if (can[x2][y]==2)
413 {
414 fg++;
415 //break;
416 }
417 for (int x=x1;x<=x2;x++)
418 if (can[x][y1]==2)
419 {
420 fg++;
421 //break;
422 }
423 if (fg==0||(fg==2&&can[x2][y1]==2&&mop[x2][y1][2][3]))
424 relax(x,i,3,dis[x.first][x.second]+1);
425 }
426 else if (x2>x1&&y2>y1)
427 {
428 int fg=0;
429 for (int y=y1;y<=y2;y++)
430 if (can[x2][y]==2)
431 {
432 fg++;
433 //break;
434 }
435 for (int x=x1;x<=x2;x++)
436 if (can[x][y1]==2)
437 {
438 fg++;
439 //break;
440 }
441 if (fg==0||(fg==2&&can[x2][y1]==2&&mop[x2][y1][2][1]))
442 relax(x,i,1,dis[x.first][x.second]+1);
443 }
444 relax(x,x.first,1,dis[x.first][x.second]+1);
445 relax(x,x.first,3,dis[x.first][x.second]+1);
446 }
447 }
448 }
449 int ans=inf;
450 ans=min(dis[cnt][0],ans);
451 ans=min(dis[cnt][1],ans);
452 ans=min(dis[cnt][2],ans);
453 ans=min(dis[cnt][3],ans);
454 if (ans==inf) puts("-1");
455 else printf("%d\n",ans);
456 }
457 return 0;
458 }View Code
Problem D
直接把180度分成3000分然后枚举取最小值即可。
#include <bits/stdc++.h>using namespace std;#define rep(i, a, b) for (int i(a); i <= (b); ++i)
#define dec(i, a, b) for (int i(a); i >= (b); --i)const double d = 3000;
const double pi = acos(-1.0);
const double g = 9.8000000;
const int N = 2003;double v[N];
double l1, r1, l2, r2;
int n;
double h;
int ans = 0;
int now;int solve(double ap){int ret = 0;rep(i, 1, n){double now;now = sqrt(v[i] * v[i] * sin(ap) * sin(ap) + 2 * h * g);now = (now - v[i] * sin(ap)) / g * v[i] * cos(ap);if (now >= l2 && now <= r2) return -1;if (now >= l1 && now <= r1) ++ret;}return ret;
} int main(){while (~scanf("%d", &n), n){ans = 0;scanf("%lf%lf%lf%lf%lf", &h, &l1, &r1, &l2, &r2);rep(i, 1, n) scanf("%lf", v + i);for (double i = -0.5 * pi; i <= 0.5 * pi; i += (pi / d)){now = solve(i);if (~now) ans = max(ans, now);}now = solve(-0.5 * pi);if (~now) ans = max(ans, now);now = solve(0);if (~now) ans = max(ans, now);now = solve(0.5 * pi);if (~now) ans = max(ans, now);printf("%d\n", ans);}return 0;
}
Problem E
Problem F
分解了十几项发现
$f_{x} = g_{a_{1}} * g_{a_{2}} * ... * g_{a_{m}}$，m为x的约数个数。
$a_{1},a_{2},...,a_{m}$分别为x的约数。
我们预处理出$g_{x}，g_{x} = f_{x} / g_{a_{1}} / g_{a_{2}} / ... / g_{a_{m-1}}$
然后就可以计算答案了。
#include <bits/stdc++.h>using namespace std;#define rep(i, a, b) for (int i(a); i <= (b); ++i)
#define dec(i, a, b) for (int i(a); i >= (b); --i)const int N = 2200;vector <int> c[N];struct dxs{int l;int d[N];dxs(){ memset(d, 0, sizeof d); l = 1;}
} a[N];int n;inline dxs chu(dxs a, dxs b){dxs ret;ret.l = a.l - b.l + 1;memset(ret.d, 0, sizeof ret.d);dec(i, a.l - 1, b.l - 1){int tt = a.d[i] / b.d[b.l - 1];if (tt != 0){rep(j, 0, b.l - 1){a.d[i - j] -= tt * b.d[b.l - 1 - j];}ret.d[i - b.l + 1] = tt;}}return ret;}bool cmp(int i, int j){if (a[i].l != a[j].l) return a[i].l < a[j].l;dec(k, a[i].l - 1, 0)if(a[i].d[k] != a[j].d[k]){if(abs(a[i].d[k]) == abs(a[j].d[k]))return a[i].d[k] < 0;else return abs(a[i].d[k]) < abs(a[j].d[k]);}return false;
}void print(dxs a){putchar('(');dec(i, a.l - 1, 0) if (a.d[i] != 0){if (i == 0){if (a.d[i] > 0) putchar('+');printf("%d", a.d[i]);}else{if (a.d[i] > 0 && i != a.l - 1) putchar('+');if (a.d[i] == -1) putchar('-');if (abs(a.d[i]) != 1) printf("%d", a.d[i]);if (i > 1) printf("x^%d", i);else putchar('x');}}putchar(')');
}int main(){rep(i, 2, 1100){rep(j, 1, i) if (i % j == 0) c[i].push_back(j);}a[1].l = 2; a[1].d[0] = -1, a[1].d[1] = 1;rep(i, 2, 1100){a[i].d[0] = -1; a[i].d[i] = 1; a[i].l = i + 1; a[i] = chu(a[i], a[1]);rep(j, 2, i - 1) if (i % j == 0) a[i] = chu(a[i], a[j]);}rep(i, 2, 1100) sort(c[i].begin(), c[i].end(), cmp);while (~scanf("%d", &n), n){if (n == 1){puts("x-1");continue;}for (auto u : c[n]) print(a[u]);putchar(10);}return 0;
}
Problem G
Problem H
先求出三维的凸包
然后枚举每个面作为底面
求出旋转后所有点的坐标
最后求出底面二维凸包的面积即可
1 #include <iostream>
2 #include <cstdio>
3 #include <cmath>
4 #include <algorithm>
5 #define eps 1e-7
6 using namespace std;
7 const double inf=0x3f3f3f3f;
8 const int MAXV=80;
9 const double EPS = 1e-9;
10 const double pi=acos(-1.0);
11 //三维点
12 struct pt {
13 double x, y, z;
14 pt() {}
15 pt(double _x, double _y, double _z): x(_x), y(_y), z(_z) {}
16 pt operator - (const pt p1) {
17 return pt(x - p1.x, y - p1.y, z - p1.z);
18 }
19 pt operator + (const pt p1) {
20 return pt(x + p1.x, y + p1.y, z + p1.z);
21 }
22 pt operator *(const pt p){
23 return pt(y*p.z-z*p.y,z*p.x-x*p.z, x*p.y-y*p.x);
24 }
25 pt operator *(double d)
26 {
27 return pt(x*d,y*d,z*d);
28 }
29
30 pt operator / (double d)
31 {
32 return pt(x/d,y/d,z/d);
33 }
34 double operator ^ (pt p) {
35 return x*p.x+y*p.y+z*p.z; //点乘
36 }
37 double len(){
38 return sqrt(x*x+y*y+z*z);
39 }
40 };
41 struct pp{
42 double x,y;
43 pp(){}
44 pp(double x,double y):x(x),y(y){}
45 };
46 struct _3DCH {
47 struct fac {
48 int a, b, c; //表示凸包一个面上三个点的编号
49 bool ok; //表示该面是否属于最终凸包中的面
50 };
51
52 int n; //初始点数
53 pt P[MAXV]; //初始点
54
55 int cnt; //凸包表面的三角形数
56 fac F[MAXV*8]; //凸包表面的三角形
57
58 int to[MAXV][MAXV];
59
60 pt Cross3(pt a,pt p){
61 return pt(a.y*p.z-a.z*p.y, a.z*p.x-a.x*p.z, a.x*p.y-a.y*p.x); //叉乘
62 }
63 double vlen(pt a) {
64 return sqrt(a.x*a.x+a.y*a.y+a.z*a.z); //向量长度
65 }
66 double area(pt a, pt b, pt c) {
67 return vlen(Cross3((b-a),(c-a))); //三角形面积*2
68 }
69 double volume(pt a, pt b, pt c, pt d) {
70 return Cross3((b-a),(c-a))^(d-a); //四面体有向体积*6
71 }
72 //三维点积
73 double Dot3( pt u, pt v )
74 {
75 return u.x * v.x + u.y * v.y + u.z * v.z;
76 }
77
78 //平面的法向量
79 pt pvec(pt a,pt b,pt c)
80 {
81 return (Cross3((a-b),(b-c)));
82 }
83 //点到面的距离
84 double Dis(pt a,pt b,pt c,pt d)
85 {
86 return fabs(pvec(a,b,c)^(d-a))/vlen(pvec(a,b,c));
87 }
88 //正：点在面同向
89 double ptof(pt &p, fac &f) {
90 pt m = P[f.b]-P[f.a], n = P[f.c]-P[f.a], t = p-P[f.a];
91 return Cross3(m , n) ^ t;
92 }
93
94 void deal(int p, int a, int b) {
95 int f = to[a][b];
96 fac add;
97 if (F[f].ok) {
98 if (ptof(P[p], F[f]) > eps)
99 dfs(p, f);
100 else {
101 add.a = b, add.b = a, add.c = p, add.ok = 1;
102 to[p][b] = to[a][p] = to[b][a] = cnt;
103 F[cnt++] = add;
104 }
105 }
106 }
107
108 void dfs(int p, int cur) {
109 F[cur].ok = 0;
110 deal(p, F[cur].b, F[cur].a);
111 deal(p, F[cur].c, F[cur].b);
112 deal(p, F[cur].a, F[cur].c);
113 }
114
115 bool same(int s, int t) {
116 pt &a = P[F[s].a], &b = P[F[s].b], &c = P[F[s].c];
117 return fabs(volume(a, b, c, P[F[t].a])) < eps && fabs(volume(a, b, c, P[F[t].b])) < eps && fabs(volume(a, b, c, P[F[t].c])) < eps;
118 }
119
120 //构建三维凸包
121 void construct() {
122 cnt = 0;
123 if (n < 4)
124 return;
125
126 /*********此段是为了保证前四个点不公面，若已保证，可去掉********/
127 bool sb = 1;
128 //使前两点不公点
129 for (int i = 1; i < n; i++) {
130 if (vlen(P[0] - P[i]) > eps) {
131 swap(P[1], P[i]);
132 sb = 0;
133 break;
134 }
135 }
136 if (sb)return;
137
138 sb = 1;
139 //使前三点不公线
140 for (int i = 2; i < n; i++) {
141 if (vlen(Cross3((P[0] - P[1]) , (P[1] - P[i]))) > eps) {
142 swap(P[2], P[i]);
143 sb = 0;
144 break;
145 }
146 }
147 if (sb)return;
148
149 sb = 1;
150 //使前四点不共面
151 for (int i = 3; i < n; i++) {
152 if (fabs(Cross3((P[0] - P[1]) , (P[1] - P[2])) ^ (P[0] - P[i])) > eps) {
153 swap(P[3], P[i]);
154 sb = 0;
155 break;
156 }
157 }
158 if (sb)return;
159 /*********此段是为了保证前四个点不公面********/
160
161
162 fac add;
163 for (int i = 0; i < 4; i++) {
164 add.a = (i+1)%4, add.b = (i+2)%4, add.c = (i+3)%4, add.ok = 1;
165 if (ptof(P[i], add) > 0)
166 swap(add.b, add.c);
167 to[add.a][add.b] = to[add.b][add.c] = to[add.c][add.a] = cnt;
168 F[cnt++] = add;
169 }
170
171 for (int i = 4; i < n; i++) {
172 for (int j = 0; j < cnt; j++) {
173 if (F[j].ok && ptof(P[i], F[j]) > eps) {
174 dfs(i, j);
175 break;
176 }
177 }
178 }
179 int tmp = cnt;
180 cnt = 0;
181 for (int i = 0; i < tmp; i++) {
182 if (F[i].ok) {
183 F[cnt++] = F[i];
184 }
185 }
186 }
187
188 //表面积
189 double area() {
190 double ret = 0.0;
191 for (int i = 0; i < cnt; i++) {
192 ret += area(P[F[i].a], P[F[i].b], P[F[i].c]);
193 }
194 return ret / 2.0;
195 }
196
197 //体积
198 double volume() {
199 pt O(0, 0, 0);
200 double ret = 0.0;
201 for (int i = 0; i < cnt; i++) {
202 ret += volume(O, P[F[i].a], P[F[i].b], P[F[i].c]);
203 }
204 return fabs(ret / 6.0);
205 }
206
207 //表面三角形数
208 int facetCnt_tri() {
209 return cnt;
210 }
211
212 //表面多边形数
213 int facetCnt() {
214 int ans = 0;
215 for (int i = 0; i < cnt; i++) {
216 bool nb = 1;
217 for (int j = 0; j < i; j++) {
218 if (same(i, j)) {
219 nb = 0;
220 break;
221 }
222 }
223 ans += nb;
224 }
225 return ans;
226 }
227 pt centroid(){
228 pt ans(0,0,0),o(0,0,0);
229 double all=0;
230 for(int i=0;i<cnt;i++)
231 {
232 double vol=volume(o,P[F[i].a],P[F[i].b],P[F[i].c]);
233 ans=ans+(o+P[F[i].a]+P[F[i].b]+P[F[i].c])/4.0*vol;
234 all+=vol;
235 }
236 ans=ans/all;
237 return ans;
238 }
239 double res(){
240 pt a=centroid();
241 double _min=1e10;
242 for(int i=0;i<cnt;++i){
243 double now=Dis(P[F[i].a],P[F[i].b],P[F[i].c],a);
244 _min=min(_min,now);
245 }
246 return _min;
247 }
248 double ptoface(pt p,int i)
249 {
250 return fabs(volume(P[F[i].a],P[F[i].b],P[F[i].c],p)/vlen((P[F[i].b]-P[F[i].a])*(P[F[i].c]-P[F[i].a])));
251 }
252 }lou;
253 bool mult(pp sp,pp ep,pp op){
254 return (sp.x-op.x)*(ep.y-op.y)>=(ep.x-op.x)*(sp.y-op.y);
255 }
256 double Cross(pp a,pp b,pp c){
257 return (c.x-a.x)*(b.y-a.y) - (b.x-a.x)*(c.y-a.y);
258 }
259
260 bool cmp(pp a,pp b){
261 if(a.y==b.y)return a.x<b.x;
262 return a.y<b.y;
263 }
264 int n,res[60],top;
265 pp ps[60];
266 void Graham(){
267 int len;
268 n=lou.n;
269 top=1;
270 sort(ps,ps+n,cmp);
271 if(n==0)return;res[0]=0;
272 if(n==1)return;res[1]=1;
273 if(n==2)return;res[2]=2;
274 for(int i=2;i<n;i++){
275 while(top&&mult(ps[i],ps[res[top]],ps[res[top-1]]))
276 top--;
277 res[++top]=i;
278 }
279 len=top;
280 res[++top]=n-2;
281 for(int i=n-3;i>=0;i--){
282 while(top!=len&&mult(ps[i],ps[res[top]],ps[res[top-1]]))top--;
283 res[++top]=i;
284 }
285 }
286 inline pt get_point(pt st,pt ed,pt tp)
287 {
288 double t1=(tp-st)^(ed-st);
289 double t2=(ed-st)^(ed-st);
290 double t=t1/t2;
291 pt ans=st + ((ed-st)*t);
292 return ans;
293 }
294 inline double dist(pt st,pt ed)
295 {
296 return sqrt((ed-st)^(ed-st));
297 }
298 pp rotate(pt st,pt ed,pt tp,double A)
299 {
300 pt root=get_point(st,ed,tp);
301 pt e=(ed-st)/dist(ed,st);
302 pt r=tp-root;
303 pt vec=e*r;
304 pt ans=r*cos(A)+vec*sin(A)+root;
305 return pp(ans.x,ans.y);
306 }
307 int main(){
308 while (scanf("%d",&lou.n)&&lou.n) {
309 for (int i=0; i<lou.n; ++i)
310 scanf("%lf%lf%lf",&lou.P[i].x,&lou.P[i].y,&lou.P[i].z);
311 lou.construct();
312 double ansh=0,ansa=inf;
313 if(lou.n<=2)
314 {
315 printf("0.000 0.000\n");
316 }
317 else if(lou.n==3)
318 {
319 ansh=0;
320 ansa=(lou.P[1]-lou.P[0])^(lou.P[2]-lou.P[0]);
321 ansa/=2.0;
322 printf("%.3lf %.3lf\n",ansh,ansa);
323
324 }
325 else {
326 for (int i=0; i<lou.cnt; ++i) {
327 pt p1=(lou.P[lou.F[i].b]-lou.P[lou.F[i].a])*(lou.P[lou.F[i].c]-lou.P[lou.F[i].a]);
328 pt e=pt(0,0,1);
329 pt vec=p1*e;
330 double A=p1^e/p1.len();
331 A=acos(A);
332 if(fabs(A-pi)>EPS&&fabs(A)>EPS){
333 pt s=pt(0,0,0);
334 for (int k=0; k<lou.n; ++k) ps[k]=rotate(s,vec,lou.P[k],A);
335 }
336 else
337 {
338 for(int k=0; k<lou.n; ++k) ps[k].x=lou.P[k].x,ps[k].y=lou.P[k].y;
339 }
340 double h=0;
341 for (int j=0; j<lou.n; ++j)
342 h=max(lou.ptoface(lou.P[j],i),h);
343 if (h<ansh)
344 continue;
345 Graham();
346 double a=0;
347 for (int k=1; k<top-1; ++k){
348 a+=fabs(Cross(ps[res[0]],ps[res[k]],ps[res[k+1]]));
349 }
350 a/=2.0;
351 if (fabs(h-ansh)<EPS) {
352 ansa=min(ansa,a);
353 }
354 else
355 ansa=a;
356 ansh=h;
357 }
358 printf("%.3lf %.3lf\n",ansh,ansa);
359 }
360 }
361 }View Code
Problem I
签到题，就是计算所有数的平方和
1 #include<iostream>
2 #include<cstdio>
3 #include<cmath>
4 #include<cstdlib>
5 #include<algorithm>
6 #include<cstring>
7 #include<string>
8 #include<vector>
9 #include<map>
10 #include<set>
11 #include<queue>
12 using namespace std;
13 typedef long long ll;
14 ll n;
15 int main()
16 {
17 while (~scanf("%lld",&n))
18 {
19 if (n==0) break;
20 ll sum=0;
21 while (n--)
22 {
23 ll x;
24 scanf("%lld",&x);
25 sum+=x*x;
26 }
27 printf("%lld\n",sum);
28 }
29 return 0;
30 }View Code
Problem J
由于只有裤子跟衣服或鞋子搭配不合法
因此可以求出每条裤子跟多少衣服和鞋子搭配是合法的
然后用乘法原理进行统计即可
1 #include<iostream>
2 #include<cstdio>
3 #include<cmath>
4 #include<cstdlib>
5 #include<algorithm>
6 #include<cstring>
7 #include<string>
8 #include<vector>
9 #include<map>
10 #include<set>
11 #include<queue>
12 using namespace std;
13 typedef long long ll;
14 int n,m,k,sum1[1010],sum2[1010];
15 char s1[11],s2[11];
16 int main()
17 {
18 while (~scanf("%d%d%d",&n,&m,&k))
19 {
20 if (n==0&&m==0&&k==0) break;
21 int q;
22 int i;
23 for (i=1;i<=m;i++)
24 {
25 sum1[i]=n;
26 sum2[i]=k;
27 }
28 scanf("%d",&q);
29 while (q--)
30 {
31 int x,y;
32 scanf("%s%d%s%d",s1,&x,s2,&y);
33 if (s1[0]=='c') sum1[y]--;
34 else sum2[x]--;
35 }
36 ll ans=0;
37 for (i=1;i<=m;i++)
38 ans+=1ll*sum1[i]*sum2[i];
39 printf("%lld\n",ans);
40 }
41 return 0;
42 }View Code
Problem K
根据题意进行模拟即可
1 #include<iostream>
2 #include<cstdio>
3 #include<cmath>
4 #include<cstdlib>
5 #include<algorithm>
6 #include<cstring>
7 #include<string>
8 #include<vector>
9 #include<map>
10 #include<set>
11 #include<queue>
12 using namespace std;
13 #define y1 dfggf
14 #define y2 kkljk
15 const int d[4][2]={{0,1},{1,0},{0,-1},{-1,0}};
16 int c1,s1,t1,c2,s2,t2,n,t;
17 int main()
18 {
19 while (~scanf("%d",&n))
20 {
21 if (n==0) break;
22 char c[2];
23 scanf("%s%d%d",c,&s1,&t1);
24 if (c[0]=='E') c1=0;
25 else if (c[0]=='S') c1=1;
26 else if (c[0]=='W') c1=2;
27 else c1=3;
28 scanf("%s%d%d",c,&s2,&t2);
29 if (c[0]=='E') c2=0;
30 else if (c[0]=='S') c2=1;
31 else if (c[0]=='W') c2=2;
32 else c2=3;
33 scanf("%d",&t);
34 int i;
35 int x1=1,y1=1,x2=n,y2=n;
36 for (i=1;i<=t;i++)
37 {
38 int nx1=x1+d[c1][0]*s1;
39 int ny1=y1+d[c1][1]*s1;
40 if (nx1<1)
41 {
42 nx1=1+(1-nx1);
43 c1=1;
44 }
45 if (nx1>n)
46 {
47 nx1=n-(nx1-n);
48 c1=3;
49 }
50 if (ny1<1)
51 {
52 ny1=1+(1-ny1);
53 c1=0;
54 }
55 if (ny1>n)
56 {
57 ny1=n-(ny1-n);
58 c1=2;
59 }
60 int nx2=x2+d[c2][0]*s2;
61 int ny2=y2+d[c2][1]*s2;
62 if (nx2<1)
63 {
64 nx2=1+(1-nx2);
65 c2=1;
66 }
67 if (nx2>n)
68 {
69 nx2=n-(nx2-n);
70 c2=3;
71 }
72 if (ny2<1)
73 {
74 ny2=1+(1-ny2);
75 c2=0;
76 }
77 if (ny2>n)
78 {
79 ny2=n-(ny2-n);
80 c2=2;
81 }
82 if (nx1==nx2&&ny1==ny2)
83 swap(c1,c2);
84 else
85 {
86 if (i%t1==0)
87 {
88 c1--;
89 if (c1==-1) c1=3;
90 }
91 if (i%t2==0)
92 {
93 //cout<<c2<<" ";
94 c2--;
95 //cout<<i<<" "<<c2<<endl;
96 if (c2==-1) c2=3;
97 }
98 }
99 x1=nx1;
100 y1=ny1;
101 x2=nx2;
102 y2=ny2;
103 //cout<<i<<endl;
104 //cout<<x1<<" "<<y1<<" "<<c1<<endl;
105 //cout<<x2<<" "<<y2<<" "<<c2<<endl;
106 }
107 printf("%d %d\n",x1,y1);
108 printf("%d %d\n",x2,y2);
109 }
110 return 0;
111 }View Code
转载于:https://www.cnblogs.com/cxhscst2/p/7626238.html
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