20200203DLUT寒假训练赛div2-简单搜索专场

:比赛地址

A - Find The Multiple

简单的dfs水题，主要是取10x和10x+1两种情况，并且注意函数存储数字必须得用unsigned long long 否则会数据溢出，出现错误

#include<cstdio>int n;bool dfs(unsigned long long int sum, int x)
{if (sum % n == 0){printf("%llu\n", sum);return 1;}if (x == 19)return 0;if (dfs(sum * 10, x + 1))return 1;if (dfs(sum * 10 + 1, x + 1))return 1;return 0;
}int main(void)
{while (~scanf("%d", &n) && n){dfs(1, 0);}
}

B - A Knight’s Journey

国际象棋的马走的规则和中国象棋一样走日，因此要枚举他能走的8个方向，但是为了让输出结果字典序最小，我们首先得让y值每次最小，其次是x，并且由于全部都能走完所以从A1点开始搜索

#include<cstdio>
#include<cstring>int p, q;
int dir[8][2] = { -1,-2,1,-2,-2,-1,2,-1,-2,1,2,1,-1,2,1,2 };
bool vis[8][8];
int steps[64];bool dfs(int x, int y, int cnt)
{if (cnt == p * q){for (int i = 0; i < cnt; ++i)printf("%c%d", steps[i] % q + 'A', steps[i] / q + 1);printf("\n");return 1;}for (int i = 0; i < 8; ++i){int tx = x + dir[i][0], ty = y + dir[i][1];if (vis[tx][ty] || tx < 0 || ty < 0 || tx >= p || ty >= q)continue;vis[tx][ty] = 1;steps[cnt] = tx * q + ty;if (dfs(tx, ty, cnt + 1))return 1;vis[tx][ty] = 0;}return 0;
}int main(void)
{int T;scanf("%d", &T);for (int i = 1; i <= T; ++i){memset(vis, 0, sizeof(vis));scanf("%d%d", &p, &q);vis[0][0] = 1;steps[0] = 0;printf("Scenario #%d:\n", i);if (!dfs(0, 0, 1))printf("impossible\n");printf("\n");}
}

C - Catch That Cow

三种走法,+1,-1,*2，枚举就可，如果初始时人已经比牛前最短的方式就是一直-1

#include<cstdio>
#include<queue>int N, K;
bool vis[100005];
struct T
{int x, step;
};
std::queue<T> q;int main(void)
{T t;scanf("%d%d", &N, &K);if (N < K){q.push(T{ N,0 });vis[N] = 1;while (!q.empty()){t = q.front();if (t.x + 1 <= 100000 && !vis[t.x + 1]){if (t.x + 1 == K){printf("%d", t.step + 1);break;}vis[t.x + 1] = 1;q.push(T{ t.x + 1,t.step + 1 });}if (t.x - 1 >= 0 && !vis[t.x - 1]){if (t.x - 1 == K){printf("%d", t.step + 1);break;}vis[t.x - 1] = 1;q.push(T{ t.x - 1 ,t.step + 1 });}if (t.x <= 50000 && !vis[t.x * 2]){if (t.x * 2 == K){printf("%d", t.step + 1);break;}vis[t.x * 2] = 1;q.push(T{ t.x * 2 ,t.step + 1 });}q.pop();}}elseprintf("%d", N - K);
}

D - Pots

利用结构体种的f来记录父亲，每次bfs拥有填充1 填充2 清空1 清空2 1倒入2 2倒入1 6种操作，并且倒入操作时考虑是否容器能够装的下，如果杯中出现了需要的水的体积就递归倒序输出，另外如果出现a||b直接等于k时可以直接一步判定，在这之外如果a==b也可以直接判定不可能

#include<cstdio>
#include<queue>
#include<cstring>
#include<cmath>int a, b, c;
bool vis[1000][1000];
struct T
{int w[2], step, c, f;
}t, q[100010];bool flag;void P(int t)
{if (q[t].f)P(q[t].f);switch (q[t].c){case 0:printf("FILL(1)\n");break;case 1:printf("FILL(2)\n");break;case 2:printf("DROP(1)\n");break;case 3:printf("DROP(2)\n");break;case 4:printf("POUR(2,1)\n");break;case 5:printf("POUR(1,2)\n");break;}
}int main(void)
{while (~scanf("%d%d%d", &a, &b, &c)){flag = 0;if (c == a)printf("1\nFILL(1)\n");else if (c == b)printf("1\nFILL(2)\n");else{if (a != b){memset(vis, 0, sizeof(vis));t.w[0] = t.w[1] = t.step = t.c = t.f = 0;vis[0][0] = 1;int r = 0, l = 0;q[r++] = t;while (l < r){t = q[l];if (!vis[a][t.w[1]]){vis[a][t.w[1]] = 1;t.w[0] = a;t.c = 0;++t.step;t.f = l;q[r++] = t;}t = q[l];if (!vis[t.w[0]][b]){vis[t.w[0]][b] = 1;t.w[1] = b;t.c = 1;++t.step;t.f = l;q[r++] = t;}t = q[l];if (!vis[0][t.w[1]]){vis[0][t.w[1]] = 1;t.w[0] = 0;t.c = 2;++t.step;t.f = l;q[r++] = t;}t = q[l];if (!vis[t.w[0]][0]){vis[t.w[0]][0] = 1;t.w[1] = 0;t.c = 3;++t.step;t.f = l;q[r++] = t;}t = q[l];if (t.w[0] + t.w[1] <= a){if (!vis[t.w[0] + t.w[1]][0]){vis[t.w[0] + t.w[1]][0] = 1;t.w[0] = t.w[0] + t.w[1];t.w[1] = 0;t.c = 4;++t.step;t.f = l;q[r++] = t;if (t.w[0] == c){flag = 1;printf("%d\n", t.step);P(r - 1);break;}}}else{if (!vis[a][t.w[0] + t.w[1] - a]){vis[a][t.w[0] + t.w[1] - a] = 1;t.w[1] = t.w[0] + t.w[1] - a;t.w[0] = a;t.c = 4;++t.step;t.f = l;q[r++] = t;if (t.w[1] == c){flag = 1;printf("%d\n", t.step);P(r - 1);break;}}}t = q[l];if (t.w[0] + t.w[1] <= b){if (!vis[0][t.w[0] + t.w[1]]){vis[0][t.w[0] + t.w[1]] = 1;t.w[1] = t.w[0] + t.w[1];t.w[0] = 0;t.c = 5;++t.step;t.f = l;q[r++] = t;if (t.w[1] == c){flag = 1;printf("%d\n", t.step);P(r - 1);break;}}}else{if (!vis[t.w[0] + t.w[1] - b][b]){vis[t.w[0] + t.w[1] - b][b] = 1;t.w[0] = t.w[0] + t.w[1] - b;t.w[1] = b;t.c = 5;++t.step;t.f = l;q[r++] = t;if (t.w[0] == c){flag = 1;printf("%d\n", t.step);P(r - 1);break;}}}++l;}}if (!flag)printf("impossible\n");}}
}

E - 非常可乐

和D题解法一致，只是多了一个瓶子而已，枚举6种倒法，且不要求打印路径，这次学会用循环写而不是针对每个特殊情况去写代码，代码量少了很多，学习ING

#include<cstdio>
#include<queue>
#include<cstring>
#include<cmath>int s[3];
bool vis[100][100][100];
struct T
{int w[3], step;
}t, q[100010];
using namespace std;
bool flag;void pour(int j, int& a, int& b)
{int cnt = a + b;b = min(cnt, s[j]);a = cnt - b;
}int main(void)
{while (~scanf("%d%d%d", &s[0], &s[1], &s[2]) && s[0] + s[1] + s[2]){flag = 0;if (s[0] & 01)printf("NO\n");else{flag = 0;memset(vis, 0, sizeof(vis));t.step = 0, t.w[0] = s[0], t.w[1] = t.w[2] = 0;int l = 0, r = 0;q[r++] = t;while (l < r && !flag){for (int i = 0; i < 3; ++i){if (q[l].w[i] > 0)for (int j = 0; j < 3; ++j){if (i == j) continue;t = q[l];pour(j, t.w[i], t.w[j]);if (!vis[t.w[0]][t.w[1]][t.w[2]]){vis[t.w[0]][t.w[1]][t.w[2]] = 1;++t.step;if (t.w[0] == s[0] / 2 && t.w[1] == s[0] / 2 || t.w[0] == s[0] / 2 && t.w[2] == s[0] / 2 || t.w[2] == s[0] / 2 && t.w[1] == s[0] / 2){flag = 1;printf("%d\n", t.step);break;}q[r++] = t;}}}++l;}if (!flag)printf("NO\n");}}
}

F - 马走日

和B题类似，甚至更简单，直接枚举就可

#include<cstdio>
#include<cstring>int p, q;
int dir[8][2] = { -1,-2,1,-2,-2,-1,2,-1,-2,1,2,1,-1,2,1,2 };
bool vis[11][11];
int ans;void dfs(int x, int y, int cnt)
{if (cnt == p * q){++ans;return;}for (int i = 0; i < 8; ++i){int tx = x + dir[i][0], ty = y + dir[i][1];if (vis[tx][ty] || tx < 0 || ty < 0 || tx >= p || ty >= q)continue;vis[tx][ty] = 1;dfs(tx, ty, cnt + 1);vis[tx][ty] = 0;}return;
}int main(void)
{int T,a,b;scanf("%d", &T);for (int i = 1; i <= T; ++i){ans = 0;memset(vis, 0, sizeof(vis));scanf("%d%d%d%d", &p, &q,&a,&b);vis[a][b] = 1;dfs(a, b, 1);printf("%d\n", ans);}
}

G - 棋盘问题

vis数组记录棋子占据的列，每次枚举从x行开始后的任意一行，如果放置的棋子数到达了k，则输出

#include<cstdio>
#include<cstring>int n, k;
bool vis[10];
char map[10][10];
int ans;void dfs(int x, int cnt)
{if (cnt == k){++ans;return;}for (int i = x; i < n; ++i)for (int j = 0; j < n; ++j){if (map[i][j] == '#' && !vis[j]){vis[j] = 1;dfs(i + 1, cnt + 1);vis[j] = 0;}}
}int main(void)
{while (~scanf("%d%d", &n, &k) && n + k > 0){ans = 0;memset(vis, 0, sizeof(vis));for (int i = 0; i < n; ++i)scanf("%s", map[i]);dfs(0, 0);printf("%d\n", ans);}}

H - 拯救行动

和一般的迷宫区别不大，就是在结构体种定义一个wait的bool变量，如果遇到守卫，wait就变为true，每次bfs的时候先检测队首是否为wait，如果是的话就把wait改为false，增加步数，再放入队列即可

#include<cstdio>
#include<cstring>
#include<queue>int m, n;
int dir[4][2] = { 0,1,1,0,0,-1,-1,0 };
bool vis[205][205];
char map[205][205];struct T
{int s, x, y;bool w;
}t;std::queue<T> q;int main(void)
{int T, sx, sy;bool flag;scanf("%d", &T);getchar();for (int i = 1; i <= T; ++i){flag = 0;sx = -1;memset(vis, 0, sizeof(vis));scanf("%d%d", &m, &n);getchar();for (int i = 0; i < m; ++i){fgets(map[i], 205, stdin);for(int j=0;map[i][j]&&sx==-1;++j)if (map[i][j] == 'r'){sx = i, sy = j;}}vis[sx][sy] = 1;t.x = sx, t.y = sy, t.w = 0, t.s = 0;q.push(t);while (!q.empty()&&!flag){t = q.front();if (!t.w){for (int i = 0; i < 4; ++i){t = q.front();int tx = t.x + dir[i][0], ty = t.y + dir[i][1];if (tx < 0 || ty < 0 || tx >= m || ty >= n || vis[tx][ty] || map[tx][ty] == '#')continue;if (map[tx][ty] == 'a'){flag = 1;printf("%d\n", t.s + 1);break;}else if (map[tx][ty] == 'x'){vis[tx][ty] = 1;t.x = tx, t.y = ty;++t.s;t.w = 1;q.push(t);}else{vis[tx][ty] = 1;t.x = tx, t.y = ty;++t.s;t.w = 0;q.push(t);}}}else{t.w = 0;++t.s;q.push(t);}q.pop();}if (flag){while (!q.empty())q.pop();}elseprintf("Impossible\n");}
}

新加的I就是在难为我，(￢︿̫̿￢☆)