POJ 2421 Constructing Roads——————最小生成树

Constructing Roads

Language:Default Constructing Roads Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 27896 Accepted: 12293 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 Source PKU Monthly,kicc

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给的邻接矩阵
和已经连接的点，让求最小生成树

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//#include<bits/stdc++.h>
#include<cstdio>
#include<vector>
#include<algorithm>

using namespace std;
const int MAXN=505;
int n,m,pre[MAXN];
struct Edge{
    int x,y,dis;
    Edge(){}
    Edge(int x1,int y1,int dis1)
    {
        x=x1; y=y1; dis=dis1;
    }
};
bool cmp(Edge a,Edge b)
{
    return a.dis<b.dis;
}
vector<Edge>d;
void init(){for(int i=1;i<=n;i++) pre[i]=i; }
int find(int x){return pre[x]==x?x:pre[x]=find(pre[x]); }
bool join(int x,int y)
{
    x=find(x);
    y=find(y);
    if(x!=y)
    {
        pre[x]=y;
        return true;
    }
    return false;
}


int main()
{

    while(~scanf("%d",&n))
    {
        d.clear();
        init();


        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
            {
                int x;
                scanf("%d",&x);
                if(i==j)                continue;
                d.push_back(Edge(i,j,x));
            }
       scanf("%d",&m);
        int a,b;
        while(m--)
        {
            scanf("%d %d",&a,&b);
            join(a,b);

        }
        sort(d.begin(),d.end(),cmp);
        int cnt=0,ans=0;
        for(int i=0;i<d.size();i++)
        {
            Edge e=d[i];
            if(join(e.x,e.y))
            {
                cnt++;
                ans+=e.dis;
            }
            if(cnt==n-1) break;
        }
        printf("%d\n",ans);
    }
    return 0;
}