#P10819. [EC Final 2020] City Brain

    ID: 10190 Type: RemoteJudge 2000ms 1024MiB Tried: 0 Accepted: 0 Difficulty: 5 Uploaded By: Tags>贪心2020三分Special JudgeO2优化最短路ICPC

[EC Final 2020] City Brain

题目描述

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 11m/s. Prof. Pang can increase the speed limit on a road by 11m/s with the cost of 11 dollar. Prof. Pang has kk dollars. He can spend any nonnegative integral amount of money on each road. If the speed limit on some road is aam/s, it takes 1/a1/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 s1s_1 to city t1t_1 and Prof. Wo starts to travel from city s2s_2 to city t2t_2. 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 s1s_1 and t1t_1 are connected by at least one path and that s2s_2 and t2t_2 are connected by at least one path.

输入格式

The first line contains three integers nn, mm, kk (1n50001\le n \le 5000, 0m50000\le m \le 5000, 0k1090\le k\le 10^9) 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 mm lines contains two integers aa, bb (1a,bn,ab1\le a, b\le n, a\neq 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 s1s_1, t1t_1, s2s_2, t2t_2 (1s1,t1,s2,t2n1\le s_1, t_1, s_2, t_2\le n) separated by single spaces denoting the starting vertices and ending vertices of Prof. Du and Prof. Wo's travels.

输出格式

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 10910^{-9}.

6 5 1
1 2
3 2
2 4
4 5
4 6
1 5 3 6
5.000000000000
1 0 100
1 1 1 1
0.000000000000
4 2 3
1 2
3 4
1 2 3 4
0.833333333333