POJ 3164 Command Network （刘朱算法|最小树形图）

Command Network

Time Limit: 1000MS		Memory Limit: 131072K
Total Submissions: 10253		Accepted: 2976

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

Source

POJ Monthly--2006.12.31, galaxy

 

http://hi.baidu.com/ggoldengoat/item/778419d01d1dcb3948e1ddb4

 

http://www.cnblogs.com/zhsl/archive/2013/02/01/2888834.html

 

解决这些：http://www.cnblogs.com/nanke/archive/2012/04/11/2441725.html

 

详细讲解：http://blog.sina.com.cn/s/blog_6af663940100ls4h.html

 

这是最小树形图的题目，1是根节点，在开始的时候自己建图。

输入n，m;代表有n个结点，接下来n行给出结点的坐标。

接下来m行给出i,j两个整数，代表i到j有连通，求出i到j的坐标距离，最后求最小树形图。

用朱刘算法可以解决。

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>

using namespace std;

const int VN=110;
const int INF=99999999;

double map[VN][VN];
int n,m,vis[VN],pt[VN][2],pre[VN];

double Cal(int i,int j){
    return sqrt(pow(pt[i][0]-pt[j][0],2.0)+pow(pt[i][1]-pt[j][1],2.0));
}

void DFS(int u){
    vis[u]=1;
    for(int i=1;i<=n;i++)
        if(!vis[i] && map[u][i]!=INF)
            DFS(i);
}

bool Connected(){   //判断是否连通，如果连通一定存在最小树形图。 
    memset(vis,0,sizeof(vis));
    DFS(1);
    for(int i=1;i<=n;i++)
        if(!vis[i])
            return 0;
    return 1;
}

double ZLEmonds(){
    int i,j,k;
    bool circle[VN];
    double ans=0;
    memset(circle,0,sizeof(circle));
    while(1){
        for(i=2;i<=n;i++){  //找出最短弧集合E0 
            if(circle[i])
                continue;
            pre[i]=i;
            map[i][i]=INF;
            for(j=1;j<=n;j++){
                if(circle[j])
                    continue;
                if(map[j][i]<map[pre[i]][i])
                    pre[i]=j;
            }
        }
        for(i=2;i<=n;i++){
            if(circle[i])
                continue;
            j=i;
            memset(vis,0,sizeof(vis));
            while(!vis[j] && j!=1){
                vis[j]=1;
                j=pre[j];
            }
            if(j==1)    //检查是否有环，能找到根节点说明没环 
                continue;
            i=j;
            ans+=map[pre[i]][i];
            for(j=pre[i];j!=i;j=pre[j]){    //i不用标记，用作后面缩点用  
                ans+=map[pre[j]][j];
                circle[j]=1;
            }
            for(j=1;j<=n;j++){
                if(circle[j])
                    continue;
                if(map[j][i]!=INF)
                    map[j][i]-=map[pre[i]][i];
            }
            for(j=pre[i];j!=i;j=pre[j]) //将环中所有的点成点i，改变边 
                for(k=1;k<=n;k++){
                    if(circle[k])
                        continue;
                    map[i][k]=min(map[i][k],map[j][k]);
                    if(map[k][j]!=INF)
                        map[k][i]=min(map[k][i],map[k][j]-map[pre[j]][j]);
                }
            break;
        }
        if(i>n){    //求出最小树形图的权值 
            for(j=2;j<=n;j++){
                if(circle[j])
                    continue;
                ans+=map[pre[j]][j];
            }
            break;
        }
    }
    return ans;
}

int main(){

    //freopen("input.txt","r",stdin);

    while(~scanf("%d%d",&n,&m)){
        int i,j;
        for(i=1;i<=n;i++)
            for(j=1;j<=n;j++)
                map[i][j]=INF;
        for(i=1;i<=n;i++)
            scanf("%d%d",&pt[i][0],&pt[i][1]);
        while(m--){
            scanf("%d%d",&i,&j);
            map[i][j]=Cal(i,j);
        }
        if(!Connected())
            printf("poor snoopy\n");
        else
            printf("%.2f\n",ZLEmonds());
    }
    return 0;
}