POJ-1258-Agri-Net

这题就是最小生成树Prim，d[]代表的是整个连接集合和未连接点的最小距离。

#include <cstdio>
int map[105][105];
int vis[105], d[105];
const int INF = 0x3f3f3f3f;
int n,ans;

void Prim()
{
    d[0] = 0;
    ans = 0;
    for (int i = 0; i < n;i++) {
        int end = -1;
        for (int j = 0; j < n;j++) {
            if (!vis[j]&&(end==-1||d[j]<d[end])) {
                end = j;
            }
        }
        vis[end] = 1;
        ans += d[end];
        for (int j = 0; j < n;j++) {
            if (!vis[j]&&d[j]>map[end][j])
                d[j] = map[end][j];
        }
    }
}

int main()
{
    while (scanf("%d",&n)!=EOF) {
        for (int i = 0; i < 105;i++) {
            vis[i] = 0;
            d[i] = INF;
        }
        for (int i = 0; i < n;i++) {
            for (int j = 0; j < n;j++) {
                scanf("%d",&map[i][j]);
            }
        }
        Prim();
        printf("%d\n", ans);
    }
    return 0;
}