Light OJ 1040 - Donation(最小生成树）

题目链接

http://www.lightoj.com/volume_showproblem.php?problem=1040

题目大意

初始有一个矩阵，代表从i到j需要花费的电缆数，现在问如果将从1到n全部连接起来，最多剩下多少电缆

解题思路

最小生成树模版题

代码如下

prime算法

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

const int N = 57;
const int INF = 0x3f3f3f3f;
int mp[N][N];
bool visited[N];
int dis[N];

int prim(int &n)
{
    memset(visited, false, sizeof(visited));
    for(int i=1; i<=n; ++ i)
        dis[i] = INF;
    dis[1] = 0;
    int res = 0;
    while(true)
    {
        int v = -1;
        for(int i=1; i<=n; ++ i)
        {
            if(visited[i] == false && (v == -1 || dis[i] < dis[v]))
                v = i;
        }
        if(v == -1)
            break;

        if(dis[v] == INF)
            return -1;

        visited[v] = true, res += dis[v];

        for(int i=1; i<=n; ++ i)
            dis[i] = min(dis[i], mp[v][i]);
    }
    return res;
}

void solve(int cases)
{
    memset(mp, 0x3f, sizeof(mp));
    int n;
    scanf("%d", &n);
    int sum = 0;
    for(int i=1; i<=n; ++ i)
        for(int j=1; j<=n; ++ j)
        {
            scanf("%d", &mp[i][j]);
            sum += mp[i][j];
            if(mp[i][j] == 0)
                mp[i][j] = INF;
            mp[i][j] = mp[j][i] = min(mp[i][j], mp[j][i]);
        }
    int res = prim(n);
    if(res == -1)
        printf("Case %d: %d\n", cases, -1);
    else
        printf("Case %d: %d\n", cases, sum - res);
}

int main()
{
    int t;
    scanf("%d", &t);
    for(int i=1; i<=t; ++ i)
        solve(i);
    return 0;
}