POJ 2421 Constructing Roads 【已知道路连通求最小生成树】

Description

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.

Input

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.

Output

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.

Sample Input

30 990 692990 0 179692 179 011 2

Sample Output

179

题目大意：给你一个N*N的邻接矩阵代表村庄1~n的连通情况， 然后t次操作， 每次操作读入两个数a b ， 代表a-b直接的道路相同； 即不需要加入最小生成树中。

    操作即把道路权值赋值为0后再查询； 这里题解多给的是克鲁斯卡尔算法， 我用的prim算法写的；

AC代码：

<strong>#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#define inf 0Xfffffffusing namespace std ;int map[2000][2000] , vis[2000] , dis[2000] , n , m ;int prim(){int mini , k , ans; for(int i = 1 ; i <=n;i++) dis[i] = map[1][i];ans = dis[1] = 0 ;for(int i = 1 ; i <=n ; i++){mini = inf ; for(int j = 1 ; j <=n ; j++){if(!vis[j]&&mini>dis[j]){mini = dis[k=j];}}vis[k] = 1 ;ans+=mini;for(int j = 1 ; j<=n; j++){if(!vis[j]&&dis[j]>map[k][j]){dis[j] = map[k][j] ;}}}return ans ; }int main(){while(cin>>n){for(int i = 0 ; i<=n;i++){for(int j = 0 ; j<=n ; j++){map[i][j] = inf ;}map[i][i] = 0 ; dis[i] = inf ;vis[i] = 0 ;}for(int i = 1 ; i<=n ; i++){for(int j = 1 ; j <=n ; j++){scanf("%d",&map[i][j]);}}int t ;cin>>t;while(t--){int a , b ; cin>>a>>b;map[a][b] = map[b][a] = 0 ;}cout<<prim()<<endl;}return 0 ;}</strong>