poj Command Network 最小树形图 朱-刘算法模板

连接：http://poj.org/problem?id=3164

Command Network

Time Limit: 1000MS		Memory Limit: 131072K
Total Submissions: 10061		Accepted: 2930

Description

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 x_i and y_i, 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 6
0 6
4 6
0 0
7 20
1 2
1 3
2 3
3 4
3 1
3 2
4 3
0 0
1 0
0 1
1 2
1 3
4 1
2 3

Sample Output

31.19
poor snoopy

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;
#define MAXN 110
#define inf 1000000000
struct Node
{
        double x;
        double y;
}node[MAXN];
int n, m;
bool visited[MAXN], circle[MAXN];
int pre[MAXN];
double graph[MAXN][MAXN];
inline double dist(int i, int j)
{
       return sqrt((node[i].x-node[j].x)*(node[i].x-node[j].x) + (node[i].y - node[j].y)*(node[i].y - node[j].y));
}
inline double min(double a, double b)
{
       if ( a < b)return a;
       return b;
}
void dfs(int u)   //深搜，判断是否存在最小树形图 ,根是否可达每个结点
{
       int i;
       if (visited[u])return;
       visited[u] = true;
       for (i = 1; i <= n; i++)if (!visited[i] && graph[u][i] != inf)dfs(i);
}
bool connect(int root)   //深搜，判断是否存在最小树形图 ,根是否可达每个结点
{
       int i;
       dfs(root);
       for (i = 1; i <= n; i++)if (!visited[i])return false;
       return true;
}
//--------------------------朱刘算法
double zhuliu(int root)
{
       int i, j, k;
       double ans = 0;
       memset(circle,0,sizeof(circle));  //如果某点被删掉，那么circle[i]=1
       while (1)
       {
  //-----------------求出除根点为root外的所有点的入边的最小值
              for (i = 1; i <= n; i++)
              {
                     if(i==root)continue;
                     if (circle[i])continue;
                     graph[i][i] = inf; //把图中所有的自环全都清除,很重要！！！
                     pre[i] = i;  //初始化自己的前一节点是自己
                     for (j = 1; j <= n; j++)  // 求i的入边的最小值
                     {
                            if (circle[j])continue;
                            if (graph[j][i] < graph[pre[i]][i])pre[i] = j;
                     }
              }
  //-------------------遍历找环
              for (i = 1; i <= n; i++)
              {
                     if(i==root)continue;
                     if (circle[i])continue;
                     j = i;
                     memset(visited,false,sizeof(visited));
                     while (!visited[j] && j != root)
                     {
                            visited[j] = true;
                            j = pre[j];
                     }
                     if (j == root)continue;//j==root说明i不在环上
                     i = j;  //找到了环
                     ans += graph[pre[i]][i];
                     for (j = pre[i]; j != i; j = pre[j])
                     {
                            ans += graph[pre[j]][j];
                            circle[j] = 1;   //用环上一点i代表此环，其他点删去，即circle[j]=1
                     }
   //判断环外的每个点A是否与环中的某个点B相连，若是相连的则改变其权值，变为graph<A,B>-graph<pre[B],B>
                     for (j = 1; j <= n; j++)
                     {
                            if (circle[j])continue;
                            if (graph[j][i] != inf)graph[j][i] -= graph[pre[i]][i];    //更新j的入边
                     }

                     for (j = pre[i]; j != i; j = pre[j])  //环上所有点的最优边为人工顶点的边
                     {
                            for (k = 1; k <= n; k++)
                            {
                                   if (circle[k])continue;
                                   if (graph[j][k] != inf)graph[i][k] = min(graph[i][k],graph[j][k]);
                                   if (graph[k][j] != inf)graph[k][i] = min(graph[k][i],graph[k][j] - graph[pre[j]][j]);
                            }
                     }
                     break;
              }
              if (i > n)
              {
                     for (j = 1; j <= n; j++)  //新图树形图的权加环的权
                     {
                            if(j==root)continue;
                            if (circle[j])continue;
                            ans += graph[pre[j]][j];
                     }
                     break;
              }

       }
       return ans;
}

int main()
{
       int i, j, u, v;
//       freopen("in.txt","r",stdin);
       while(scanf("%d %d",&n,&m) != EOF)
       {
              for (i = 1; i <= n; i++)scanf("%lf %lf",&node[i].x,&node[i].y);
              for (i = 1; i <= n; i++)for (j = 1; j <= n; j++)graph[i][j] = inf;
              while (m--)
              {
                     scanf("%d %d",&u,&v);
                     graph[u][v] = dist(u,v);
              }
              memset(visited,false,sizeof(visited));
              int root=1;
              if (!connect(root))printf("poor snoopy\n");
              else printf("%.2lf\n",zhuliu(root));
       }
       return 0;
}