0
点赞
收藏
分享

微信扫一扫

Constructing Roads 最小生成树 消除已经建过的路的影响

J简文 2022-02-20 阅读 47
算法c++

题目中两个村庄如果没有路的话长度为0,为了消除已经建过的路对结果的影响,将没有路的长度设置为-1,已经建过路的村庄之间的路长度为0
然后用prime算法求出最小生成树即可

#include <cstring>
#include <iostream>
#define mem(a, v) memset(a, v, sizeof(a))
#define fre(f) freopen(f ".in", "r", stdin), freopen(f ".out", "w", stdout)
#define inf 0x3f3f3f3f
const int N = 1e2 + 5;
using namespace std;

int g[N][N], vis[N], sum = 0;

int main()
{
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    mem(vis, 0);
    int n, m, x, y;
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= n; j++)
        {
            cin >> g[i][j];
            if (g[i][j] == 0) g[i][j] = -1; //没有路,长度为-1
        }
    }
    cin >> m;
    for (int i = 1; i <= m; i++)
    {
        cin >> x >> y;
        g[x][y] = g[y][x] = 0; //建过路了就将长度设置为0,加入了树中对结果也不会有影响
    }
    vis[1] = 1;
    for (int i = 1; i < n; i++)
    {
        int minlen = inf, ans;
        for (int j = 1; j <= n; j++)
        {
            if (!vis[j]) continue;
            for (int k = 1; k <= n; k++)
            {
                if (vis[k] || k == j || g[j][k] == -1) continue;
                if (g[j][k] < minlen)
                {
                    minlen = g[j][k];
                    ans = k;
                }
            }
        }
        vis[ans] = 1;
        sum += minlen;
    }
    cout << sum;
    return 0;
}
举报

相关推荐

0 条评论