hdu1102——Constructing Roads（prim）

Problem 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
3
0 990 692
990 0 179
692 179 0
1
1 2

Sample Output
179

给出点与点之间的距离，并且再给出一些两个点之间已经修好了路，求最少还需修哪些路使得每个点之间连通
最小生成树，已修的点之间距离变成0，再用prim算法过一下。
题目是多组输入数据

#include <iostream>#include <cstring>#include <string>#include <vector>#include <queue>#include <cstdio>#include <set>#include <math.h>#include <algorithm>#include <queue>#include <iomanip>#define INF 0x3f3f3f3f#define MAXN 200#define Mod 99999999using namespace std;int cost[MAXN][MAXN],vis[MAXN],lowc[MAXN],n;int prim(){    int ans=0;    memset(vis,0,sizeof(vis));    vis[0]=1;    for(int i=1; i<n; ++i)        lowc[i]=cost[0][i];    for(int i=1; i<n; ++i)    {        int minc=INF;        int p=-1;        for(int j=0; j<n; ++j)            if(!vis[j]&&minc>lowc[j])            {                minc=lowc[j];                p=j;            }        if(minc==INF)            return -1;        ans+=minc;        vis[p]=1;        for(int j=0; j<n; ++j)            if(!vis[j]&&lowc[j]>cost[p][j])                lowc[j]=cost[p][j];    }    return ans;}int main(){    while(~scanf("%d",&n))    {        for(int i=0; i<n; ++i)            for(int j=0; j<n; ++j)                scanf("%d",&cost[i][j]);        int q;        scanf("%d",&q);        while(q--)        {            int a,b;            scanf("%d%d",&a,&b);            cost[a-1][b-1]=0;            cost[b-1][a-1]=0;        }        printf("%d\n",prim());    }    return 0;}