2-2 畅通工程之局部最小花费问题 (30 分)【普利姆算法】

4
1 2 1 1
1 3 4 0
1 4 1 1
2 3 3 0
2 4 2 1
3 4 5 0

3

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define inf 999999999
int n,e[101][101],vis[101],dis[101];
int main(){
    int a,b,statue,price;
    scanf("%d",&n);
    for(int i = 1;i <= n;i ++){
        for(int j = 1;j <= n;j ++)e[i][j] = e[j][i] = inf;
        e[i][i] = 0;
    }//初始化 
    for(int i = 1;i <= n*(n-1)/2;i ++){
        scanf("%d%d%d%d",&a,&b,&price,&statue);
        if(statue)e[a][b] = e[b][a] = 0;
        else e[a][b] = e[b][a] = price;//存值 
    }
    vis[1] = 1;//判断是否访问过 
    for(int i = 1;i <= n;i ++)dis[i] = e[1][i];//存值普利姆算法 
    int count = 1;//判断是否访问完 
    int sum = 0;//输出最小值 
    while(count < n){
        int min = inf,d = 0;
        for(int i = 1;i <= n;i ++)if(vis[i] == 0&&dis[i] < min)min = dis[i],d = i;//更新最小 
        vis[d] = 1;count ++;sum += min;//更新值 
        for(int i = 1;i <= n;i ++)if(vis[i] == 0&&dis[i]>e[d][i])dis[i] = e[d][i];//更新最小值 
    }printf("%d",sum);
}