Prim

n^2f复杂度

生成单点

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
#include<math.h>
using namespace std;
int n,way[120][120],minn[120],t[120],ans=0;
int main()
{
    freopen("WIRE.IN","r",stdin);
    freopen("WIRE.OUT","w",stdout);
    int x;    
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    for(int j=1;j<=n;j++)    
        {
            scanf("%d",&x);
            way[i][j]=x;
        }
    memset(minn,127,sizeof(minn));
    minn[1]=0;
    for(int i=1;i<=n;i++)
    {
        int k=0;
        for(int j=1;j<=n;j++)
        if((!t[j])&&minn[j]<minn[k])
            k=j;
        t[k]=1;
        for(int j=1;j<=n;j++)
        if(!t[j])    minn[j]=min(minn[j],way[k][j]);        
    }
    for(int i=1;i<=n;i++)
    ans+=minn[i];
    cout<<ans;
    return 0;
} 
原文地址:https://www.cnblogs.com/CLGYPYJ/p/6385023.html