最小树形图 Command Network POJ - 3164

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本文介绍了一种基于图论的算法——朱刘算法，并提供了一个详细的实现模板。该算法主要用于解决带方向边的最小生成树问题，适用于竞赛编程中涉及图论的部分题目。通过实例演示了如何计算顶点之间的距离并构建最小生成树。

朱刘算法模版

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <cmath>
#include <vector>

using namespace std;

struct node
{
    int u, v, next;
    double w;
}e[20001];

double x[101], y[101];
int n, m;

double directedmst()
{
    int root = 1;
    double ans = 0;
    double in[101];
    int id[101], pre[101], vis[101];

    for (;;)
    {
        for (int i = 1; i <= n; i++)
            in[i] = 1e9;
        for (int i = 1; i <= m; i++)
            if (e[i].w < in[e[i].v] && e[i].u != e[i].v)
                pre[e[i].v] = e[i].u, in[e[i].v] = e[i].w;
        for (int i = 1; i <= n; i++)
        {
            if (i == root)
                continue;
            if (in[i] == 1e9)
                return -1;
        }

        int cnt = 0;
        memset(id, -1, sizeof(id));
        memset(vis, -1, sizeof(vis));
        in[root] = 0;
        for (int i = 1; i <= n; i++)
        {
            ans += in[i];
            int v = i;
            while (vis[v] != i && id[v] == -1 && v != root)
            {
                vis[v] = i;
                v = pre[v];
            }
            if (root != v && id[v] == -1)
            {
                cnt++;
                for (int u = pre[v]; u != v; u = pre[u])
                    id[u] = cnt;
                id[v] = cnt;
            }
        }
        if (cnt == 0)
            break;
        for (int i = 1; i <= n; i++)
            if (id[i] == -1)
                id[i] = ++cnt;
        for (int i = 1; i <= m; i++)
        {
            int u = e[i].u, v = e[i].v;
            e[i].u = id[u];
            e[i].v = id[v];
            if (id[u] != id[v])
                e[i].w -= in[v];
        }
        n = cnt;
        root = id[root];
    }

    return ans;
}

int main()
{
    while (~scanf("%d %d", &n, &m))
    {
        for (int i = 1; i <= n; i++)
            scanf("%lf %lf", &x[i], &y[i]);
        for (int i = 1; i <= m; i++)
        {
            scanf("%d %d", &e[i].u, &e[i].v);
            e[i].w = sqrt((x[e[i].u] - x[e[i].v]) * (x[e[i].u] - x[e[i].v]) + (y[e[i].u] - y[e[i].v]) * (y[e[i].u] - y[e[i].v]));
            if (e[i].u == e[i].v)
                e[i].w = 1e9;
        }
        double ans = directedmst();
        if (ans == -1)
            printf("poor snoopy\n");
        else
            printf("%.2lf\n", ans);
    }

    return 0;
}