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There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a village C such that there is a road between A and C, and C and B are connected.
We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum.
The first line is an integer N (3 <= N <= 100), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 1000]) between village i and village j.
Then there is an integer Q (0 <= Q <= N * (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1 <= a < b <= N), which means the road between village a and village b has been built.
You should output a line contains an integer, which is the length of all the roads to be built such that all the villages are connected, and this value is minimum.
3 0 990 692 990 0 179 692 179 0 1 1 2
179
有N个村庄,用公路连接起来(间接连通也可以)。给出一个矩阵代表两个村庄之间的距离,其中已经有一些村庄连起来了,问最少还要建多长的路才能把所有点连起来;典型的最小生成树的模板题。就不再贴出代码了。
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原文地址:http://www.cnblogs.com/liuzhanshan/p/5648491.html