EOJ 3199/POJ 1258 Agri-Net【Prim】

題目簡介

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好像就是直接求最小生成樹啊……

說明

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就是記錄一下寫得比較簡潔的Prim算法而已。

#include <bits/stdc++.h>
using namespace std;

const int maxn = 105;
const int INF = 0x3f3f3f3f;
int mp[maxn][maxn], dis[maxn], vis[maxn], n, sum;

inline int prim()
{
    memset(vis, 0, sizeof vis);
    memset(dis, INF, sizeof dis);
    sum = dis[1] = 0;
    for (;;)
    {
        int u = -1, v, mincost = INF;
        for (int i = 1; i <= n; ++i)
            if (dis[i] < mincost && !vis[i])
            {
                mincost = dis[i];
                u = i;
            }
        if (u == -1) break;
        vis[u] = 1;
        sum += dis[u];
        for (v = 1; v <= n; ++v)
            if (dis[v] > mp[u][v]) dis[v] = mp[u][v];
    }
    return sum;
}

int main()
{
    while (cin >> n)
    {
        memset(mp, INF, sizeof mp);
        for (int i = 1; i <= n; ++i)
            for (int j = 1; j <= n; ++j)
                scanf("%d", &mp[i][j]);
        printf("%d\n", prim());
    }
    return 0;
}