2020ICPC EC-Final D. City Brain

题目链接
Prof. Pang works for the City Brain program of Capital Grancel. The road network of Grancel can be represented by an undirected graph. Initially, the speed limit on each road is 1 1 1m/s. Prof. Pang can increase the speed limit on a road by 1 1 1m/s with the cost of 1 1 1 dollar. Prof. Pang has k k k dollars. He can spend any nonnegative integral amount of money on each road. If the speed limit on some road is a a am/s, it takes 1 / a 1/a 1/a seconds for anyone to go through the road in either direction.

After Prof. Pang spent his money, Prof. Du starts to travel from city s 1 s_1 s1 to city t 1 t_1 t1 and Prof. Wo starts to travel from city s 2 s_2 s2 to city t 2 t_2 t2. Help Prof. Pang to spend his money wisely to minimize the sum of minimum time of Prof. Du’s travel and Prof. Wo’s travel. It is guaranteed that s 1 s_1 s1 and t 1 t_1 t1 are connected by at least one path and that s 2 s_2 s2 and t 2 t_2 t2 are connected by at least one path.

Input

The first line contains three integers n n n, m m m, k k k ( 1 ≤ n ≤ 5000 1\le n \le 5000 1≤n≤5000, 0 ≤ m ≤ 5000 0\le m \le 5000 0≤m≤5000, 0 ≤ k ≤ 1 0 9 0\le k\le 10^9 0≤k≤109) separated by single spaces denoting the number of vertices, the number of edges in the graph and the number of dollars Prof. Pang has.

Each of the following m m m lines contains two integers a a a, b b b ( 1 ≤ a , b ≤ n , a ≠ b 1\le a, b\le n, a\neq b 1≤a,b≤n,a=b) separated by a single space denoting the two endpoints of one road. There can be multiple roads between the same pair of cities.

The following line contains four integers s 1 s_1 s1, t 1 t_1 t1, s 2 s_2 s2, t 2 t_2 t2 ( 1 ≤ s 1 , t 1 , s 2 , t 2 ≤ n 1\le s_1, t_1, s_2, t_2\le n 1≤s1,t1,s2,t2≤n) separated by single spaces denoting the starting vertices and ending vertices of Prof. Du and Prof. Wo’s travels.

Output

Output one decimal in the only line – the minimum sum of Prof. Du’s travel time and Prof. Wo’s travel time. The answer will be considered correct if its absolute or relative error does not exceed 1 0 − 9 10^{-9} 10−9.

Examples

Input

Output

5.000000000000

显然根据贪心原则，要把修改次数用完。
对于长度为 l l l 的路径，假设有 x x x 次机会修改，则根据贪心原则，修改要尽可能平均，此时通过该路径的用时为 x m o d l ⌊ x l ⌋ + 2 + l − ( x m o d l ) ⌊ x l ⌋ + 1 \frac{x\mod l}{\left \lfloor \frac{x}{l} \right \rfloor+2}+\frac{l-(x\mod l)}{\left \lfloor \frac{x}{l} \right \rfloor+1} ⌊lx⌋+2xmodl+⌊lx⌋+1l−(xmodl) ,这个可以 O ( 1 ) O(1) O(1) 计算。路径可以分为重复和不重复两部分，假设不重复部分分配 x x x 次修改，则重复部分分配 k − x k-x k−x 次修改。感性理解一下：由于重复部分和不重复部分用时关于 x x x 的函数的导数均为减函数，因此总用时关于 x x x 的函数为单峰函数或单调函数，利用三分可以 O ( log ⁡ k ) O(\log k) O(logk) 时间求出最值。
对于路径，首先可以观察到图为稀疏图且边权相等，因此可以通过枚举起点 bfs O ( n 2 ) O(n^2) O(n2) 预处理出任意两点的最短路。之后 O ( n 2 ) O(n^2) O(n2) 枚举重复路径的起点和终点得到重复路径和不重复路径长度，然而直接三分 O ( n 2 log ⁡ k ) O(n^2\log k) O(n2logk) 会超时，因此先筛掉相同重复路径长度中不重复路径长度较大的方案，然后枚举重复路径长度就可以将复杂度降为 O ( n log ⁡ k ) O(n\log k) O(nlogk) 。
最终时间复杂度为 O ( n 2 + n log ⁡ k ) O(n^2+n\log k) O(n2+nlogk) 。

#include <bits/stdc++.h>#define get(l, x) ((l) ? (double((x) % (l)) / double((x) / (l) + 2) + double((l) - (x) % (l)) / double((x) / (l) + 1)) : 0)
#define f(l1, l2, x) (get(l1, x) + 2 * get(l2, k - (x)))
using namespace std;
const int N = 5005, INF = 0x3f3f3f3f;
int head[N], ver[N << 1], Next[N << 1], tot;
int n, m, k, s1, t1, s2, t2, d[N][N], mn[N];
queue<int> q;inline void add(int x, int y) {ver[++tot] = y;Next[tot] = head[x];head[x] = tot;
}inline void bfs(int s) {memset(d[s] + 1, 0x3f, sizeof(int) * n);q.push(s), d[s][s] = 0;while (!q.empty()) {int x = q.front();q.pop();for (int i = head[x]; i; i = Next[i]) {if (d[s][ver[i]] != INF) continue;d[s][ver[i]] = d[s][x] + 1, q.push(ver[i]);}}
}inline double calc(int l1, int l2) {if (!l1 && !l2) return 0;if (!l2) return get(l1, k);if (!l1) return 2 * get(l2, k);int l = 0, r = k, lmid, rmid, ans;while (l <= r) {lmid = l + (r - l) / 3, rmid = r - (r - l) / 3;f(l1, l2, lmid) >= f(l1, l2, rmid) ? (l = lmid + 1, ans = rmid) : (r = rmid - 1);}return f(l1, l2, ans);
}int main() {scanf("%d%d%d", &n, &m, &k);for (int i = 1, x, y; i <= m; i++) {scanf("%d%d", &x, &y);add(x, y), add(y, x);}for (int i = 1; i <= n; i++) bfs(i);scanf("%d%d%d%d", &s1, &t1, &s2, &t2);memset(mn + 1, 0x3f, sizeof(int) * n);mn[0] = d[s1][t1] + d[s2][t2];for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++) {if (d[i][j] == INF || d[s1][i] == INF || d[s2][i] == INF || d[j][t1] == INF || d[j][t2] == INF) continue;mn[d[i][j]] = min(mn[d[i][j]], min(d[s1][i] + d[j][t1], d[s1][j] + d[i][t1]) + min(d[s2][i] + d[j][t2], d[s2][j] + d[i][t2]));}double ans = 1e9;for (int i = 0; i < n; i++) {if (mn[i] == INF) continue;ans = min(ans, calc(mn[i], i));}printf("%.15lf\n", ans);return 0;
}

2020ICPC EC-Final D. City Brain相关推荐

2020 ICPC Asia East Continent Final D. City Brain(最短路+三分)
传送门题意: 给出nnn 个点,mmm条边的无向带权图,初始边权都为111,一共有kkk 次操作机会,每次操作可以选择一条边使其边权+1+1+1, 通过一条边的时间为 1/1/1/边权 ,求mi ...
City Brain
题目链接:City Brain 显然两条路径之间,相交的部分一定是一段连续的路径. 所以我们可以枚举连续的路径,然后剩下4段不连续的路径.我们可以预处理两点之间的最短路,然后算出分别的和,然后三分即可 ...
2020 EC Final 诸事不顺记
好像又快过了一星期了...简要记一下好了. day -14 昆明之后的两个星期,第一个星期在搞 GDOI 各项准备工作,第二个星期在肝编译原理大作业.编译原理理论作业.通信原理作业. 就这 ...
2020 ICPC EC Final 自闭记
我搬运我自己应该算原创吧前些天教练突然多拿到一个EC的名额, 然后给了我们qaq, 代价是沈阳不能去. Day -1 Day 1 Day 2 Day -1 我们五点多还在写台账, 张老师就带cdcq ...
组合计数 ---- 2020 EC final B. Rectangle Flip 2(枚举+组合计数)
题目大意题目大意: 给你一个n×mn\times mn×m的矩阵问你可以截取出多少个不同的矩阵然后每次会有一个点变成无效点动态输出每次剩余可挖的矩阵是多少? 解题思路: 首先我们可以这么搞就是对 ...
2018年EC Final 校内选拔赛【解题报告】
问题 A: C基础-求同存异时间限制: 1 Sec 内存限制: 128 MB 提交: 1 解决: 1 [提交][状态][讨论版][命题人:外部导入][Edit] [TestData] 题目描述 ...
2018 EC Final（西安）小结
2018最后一场+队伍最后一场现场赛,银热身赛 3道题,第一题签到,第二题是一道平面几何,不难,但是貌似因为一些玄学问题(精度?),不少人都WA了好几次,但最后还是过了,第三题是输入两个数输出一个数 ...
【LOJ6713】「EC Final 2019」狄利克雷 k 次根加强版（狄利克雷生成函数）
传送门题解: 我记得 SCOI2019 考完之后我就口胡过这个东西,当时D1T3 超矩形... 考虑 Dirichlet 生成函数:F(x)=∑i≥1fiixF(x)=\sum_{i\geq 1}\ ...
[Ec Final 2018] Misunderstood … Missing
题目描述: 我有两个基本属性一个 A 表示每次攻击可以造成的伤害,D为每轮自动增加的攻击力(A D初始值都为0 每轮你有三种操作(任选一个 1:造成 A+ai 的伤害 2:给 D 增加 Bi 的数值 ...

2020ICPC EC-Final D. City Brain

2020ICPC EC-Final D. City Brain相关推荐

最新文章

热门文章