标签：最小生成树

继续畅通工程

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

3
1
0

题意：求最少的修路价格

题解：Prime算法 最小生成树 从1开始  感觉不是很难 懒得多说了就 直接上代码

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

const int maxn = 111;
const int inf = 0x3f3f3f3f;
int gra[maxn][maxn];
int dist[maxn];
bool vis[maxn];
int n;

int prime();

int main(){
    while(scanf("%d", &n)!=EOF && n){

        memset(gra, inf, sizeof(gra));
        memset(vis, false, sizeof(vis));

        int a, b, cost, state;

        for(int i = 0; i < n*(n-1)/2; ++i){
            scanf("%d%d%d%d", &a, &b, &cost, &state);
            if(state == 0){
                gra[a][b] = gra[b][a] = cost;
            }
            else{
                gra[a][b] = gra[b][a] = 0;
            }
        }

        printf("%d\n", prime());

    }
    return 0;
}

int prime(){
    int sum = 0;
    int mi, index;

    for(int i = 1; i <= n; ++i){
        dist[i] = gra[1][i];
    }

    vis[1] = true;
    dist[1] = inf;

    for(int i = 1; i < n; ++i){
        mi = inf;
        for(int j = 1; j <= n; ++j){
            if(!vis[j] && dist[j] < mi){
                mi = dist[j];
                index = j;
            }
        }

        sum += dist[index];
        vis[index] = true;

        for(int j = 1; j <= n; ++j){
            if(!vis[j] && dist[j] > gra[index][j]){
                dist[j] = gra[index][j];
            }
        }
    }
    return sum;
}

【POJ】1879继续畅通工程

标签：最小生成树

原文地址：http://blog.csdn.net/u012431590/article/details/45974535