本文共 4010 字，大约阅读时间需要 13 分钟。

题意：

1到所有点的最短距离，有向，最短，最小树形图（朱刘算法），板子题。没有任何坑点。

不会的点。 学吧！ 挺简单的。

注意：

别交G++，用C++交

原因如下：   After a long lasting war on words, a war on arms finally breaks out between littleken’s and KnuthOcean’s kingdoms. A sudden and violent assault by KnuthOcean’s force has rendered a total failure of littleken’s command network. A provisional network must be built immediately. littleken orders snoopy to take charge of the project.

  With the situation studied to every detail, snoopy believes that the most urgent point is to enable littenken’s commands to reach every disconnected node in the destroyed network and decides on a plan to build a unidirectional communication network. The nodes are distributed on a plane. If littleken’s commands are to be able to be delivered directly from a node A to another node B, a wire will have to be built along the straight line segment connecting the two nodes. Since it’s in wartime, not between all pairs of nodes can wires be built. snoopy wants the plan to require the shortest total length of wires so that the construction can be done very soon.

Input

  The input contains several test cases. Each test case starts with a line containing two integer N (N ≤ 100), the number of nodes in the destroyed network, and M (M ≤ 104), the number of pairs of nodes between which a wire can be built. The next N lines each contain an ordered pair xi and yi, giving the Cartesian coordinates of the nodes. Then follow M lines each containing two integers i and j between 1 and N (inclusive) meaning a wire can be built between node i and node j for unidirectional command delivery from the former to the latter. littleken’s headquarter is always located at node 1. Process to end of file.

Output

  For each test case, output exactly one line containing the shortest total length of wires to two digits past the decimal point. In the cases that such a network does not exist, just output ‘poor snoopy’.

Sample Input

4 60 64 60 07 201 21 32 33 43 13 24 30 01 00 11 21 34 12 3

Sample Output

31.19poor snoopy

代码：

#include 
   
    #include 
    
     int n, m;double flag;struct node {       int be, en; double va;}a[10010];double zhuliu(int root) {       int  pre[101], consume[101], vis[101], i;    double sum = 0, dis[101];    while (true) {           for (i = 1; i <= n; i++)dis[i] = flag;        for (i = 1; i <= m; i++)            if (a[i].be != a[i].en && a[i].va < dis[a[i].en])                pre[a[i].en] = a[i].be, dis[a[i].en] = a[i].va;        for (i = 1; i <= n; i++) {               if (i != root && dis[i] == flag)return -1;            vis[i] = consume[i] = 0;        }        int cnt = 0;        for (i = 1; i <= n; i++) {               if (i == root)continue;            sum += dis[i];            int v = i;            while (vis[v] != i && v != root && !consume[v])                vis[v] = i, v = pre[v];            if (!consume[v] && v != root) {                   consume[v] = ++cnt;                for (int u = pre[v]; u != v; u = pre[u])                    consume[u] = cnt;            }        }        if (!cnt)return sum;        for (i = 1; i <= n; i++)            if (!consume[i])consume[i] = ++cnt;        for (i = 1; i <= m; i++) {               int u = a[i].be;            int v = a[i].en;            a[i].be = consume[u];            a[i].en = consume[v];            if (consume[u] != consume[v])a[i].va -= dis[v];        }        root = consume[root]; n = cnt;    }}double calculate(double a, double b, double c, double d) {       return sqrt((a - b) * (a - b) + (c - d) * (c - d));}int main() {       while (~scanf("%d %d", &n, &m)) {           double x[101], y[101]; flag = 1; int i;        for (i = 1; i <= n; i++)scanf("%lf %lf", &x[i], &y[i]);        for (i = 1; i <= m; i++) {               scanf("%d %d", &a[i].be, &a[i].en);            a[i].va = calculate(x[a[i].be], x[a[i].en], y[a[i].be], y[a[i].en]);            flag += a[i].va;        }        double re = zhuliu(1);        if (re == -1)printf("poor snoopy\n");        else printf("%.2lf\n", re);    }}
    
   

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