題目大意：有N個點，M條有向邊。

如今要求你以1為根。構造出一棵最小生成樹，問這棵最小生成樹是否能被構造出來，假設能夠。總權值是多少

解題思路：朱劉算法的裸題。我僅僅想吐槽一下POJ，用的是double型的，輸出時是%.2lf，結果是WA
換成了%.2f就A了。

。這什麽情況，白白花費了1個多小時去調錯。。

#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;

#define N 110
#define M 10010
#define esp 1e-5
 

const double INF = 1e9;

struct Node {
    int x, y;
}node[N];

struct Edge{
    int from, to;
    double dis;
}E[M];

int n, m, tot;
int pre[N], id[N], vis[N];
double in[N];

double distance(int u, int v) {
    int x = node[u].x - node[v].x;
    int y = node[u].y - node[v].y;

    return sqrt(1.0 * x * x + 1.0
 
 * y * y);
}

void AddEdge(int u, int v) {
    E[tot].from = u;
    E[tot].to = v;
    E[tot].dis = distance(u, v);
    tot++;
}

void init() {
    for (int i = 0; i < n; i++) 
        scanf("%d%d", &node[i].x, &node[i].y);

    tot = 0;
    int u, v;
    for (int i = 0; i < m; i++) {
        scanf("%d%d"
 
, &u, &v);
        u--; v--;
        AddEdge(u, v);
    }
}

double ans;
bool Directed_MST(int root) {
    ans = 0;
    int u, v, tmp;
    while (1) {
        for (int i = 0; i < n; i++)
            in[i] = INF;

        for (int i = 0; i < m; i++) {
            u = E[i].from;
            v = E[i].to;
            if (u != v && E[i].dis - in[v] < esp) {
                in[v] = E[i].dis;
                pre[v] = u;
            }
        }

        for (int i = 0; i < n; i++) {
            if (i == root)
                continue;
            if (in[i] == INF)
                return false;
        }

        int subnode = 0;
        memset(vis, -1, sizeof(vis));
        memset(id, -1, sizeof(id));
        in[root] = 0;

        for (int i = 0; i < n; i++) {
            ans += in[i];
            tmp = i;
            while (vis[tmp] != i && id[tmp] == -1 && tmp != root) {
                vis[tmp] = i;
                tmp = pre[tmp];
            }

            if (tmp != root && id[tmp] == -1) {
                u = pre[tmp];
                while (u != tmp) {
                    id[u] = subnode;
                    u = pre[u];
                }
                id[tmp] = subnode++;
            }
        }

        if (subnode == 0)
            return true;

        for (int i = 0; i < n; i++)
            if (id[i] == -1)
                id[i] = subnode++;

        for (int i = 0; i < m; i++) {
            tmp = E[i].to;
            E[i].from = id[E[i].from];
            E[i].to = id[E[i].to];

            if (E[i].from != E[i].to) 
                E[i].dis -= in[tmp];
        }

        root = id[root];
        n = subnode;
    }
}

void solve() {
    if (!Directed_MST(0)) {
        printf("poor snoopy\n");
    }
    else 
        printf("%.2f\n", ans);
}

int main() {
    while (~scanf("%d%d", &n, &m)) {
        init();
        solve();
    }
    return 0;
}

POJ - 3164 Command Network(朱劉算法)