Constructing Roads

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 31254    Accepted Submission(s): 11800

 

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

AC代码:

#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;
const int MAX = 105;
int G[MAX][MAX], ans = 0;
struct node{
    int u, v, w;
    bool operator < (const node a) const{
        return w < a.w;
    }
};
vector<node> V;
int Pre[MAX];

int find(int a){
    int c = a;
    while(a != Pre[a])
        a = Pre[a];
    Pre[c] = a;
    return a;
}
bool join(int u, int v){
    int fu = find(u), fv = find(v);
    if(fu != fv){
        Pre[fv] = fu;
        return true;
    }
    return false;
}
void solve(){
    for(int i = 0; i < V.size(); i++)
        //利用并查集,如果连通成功,这个边的权值计入答案
        if(join(V[i].u, V[i].v)) ans += V[i].w;
    printf("%d\n",ans);
}
int main(){
    int n;
    while(scanf("%d",&n)!=EOF){
        //初始化
        V.clear();
        ans=0;
        for(int i = 1; i <= n; i++) Pre[i] = i;
        //建图
        for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j++)
            scanf("%d",&(G[i][j]));
        int k;scanf("%d",&k);
        while(k--){//把已经连同的村庄 w 设置为0
            int x,y;scanf("%d%d",&x,&y);
            G[x][y] = G[y][x] = 0;
        }
        //图中很多数据多余,只要右上三角的值就可以
        for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j++){
            if(j > i){
                node a;
                a.u = i,a.v = j, a.w = G[i][j];
                V.push_back(a);
            }
        }
        //Kruskal算法,按边的权值排序,然后依次连同
        sort(V.begin(), V.end());
        solve();
    }
    return 0;
}