POJ 3164 Command Network 最小树形图模板题

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探讨在战争条件下如何快速建立一个单向通信的指挥网络，确保命令能够从总部传达至每一个节点，同时考虑到实际情况限制，寻求最短总连线长度。

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Command Network

Time Limit: 1000MS		Memory Limit: 131072K
Total Submissions: 10899		Accepted: 3184

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

Sample Output

31.19
poor snoopy

Source

POJ Monthly--2006.12.31, galaxy

题意是说部队之间电缆传达信息，传达方向是单向的，现在给出部队位置和可以连接的部队，问最少需要多长的电缆。

最基础的最小树形图问题。

#include<stdio.h>
#include<string.h>
#include<math.h>
#define M 107
#define inf 0x3f3f3f
using namespace std;
int pre[M],vis[M],id[M];
double in[M];
int n,m;

struct Node
{
    int u,v;
    double cost;
}E[M*M+5];

struct point
{
    double x,y;
}p[M];

double dis(point a,point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

double direct_mst(int root,int nv,int ne)
{
    double ret=0;
    while(true)
    {
        //找最小入边
        for(int i=0;i<nv;i++)
            in[i]=inf;
        for(int i=0;i<ne;i++)
        {
            int u=E[i].u;
            int v=E[i].v;
            if(E[i].cost<in[v]&&u!=v)
            {
                pre[v]=u;
                in[v]=E[i].cost;
            }
        }
        for(int i=0;i<nv;i++)
        {
            if(i==root)continue;
            if(in[i]==inf)return -1;
        }
        //找环
        int cntnode=0;
        memset(id,-1,sizeof(id));
        memset(vis,-1,sizeof(vis));
        in[root]=0;
        for(int i=0;i<nv;i++)//标记每个环
        {
            ret+=in[i];
            int v=i;
            while(vis[v]!=i&&id[v]==-1&&v!=root)
            {
                vis[v]=i;
                v=pre[v];
            }
            if(v!=root&&id[v]==-1)
            {
                for(int u=pre[v];u!=v;u=pre[u])
                {
                    id[u]=cntnode;
                }
                id[v]=cntnode++;
            }
        }
        if(cntnode==0)break;//无环
        for(int i=0;i<nv;i++)
            if(id[i]==-1)
                id[i]=cntnode++;
        //缩点，重新标记
        for(int i=0;i<ne;i++)
        {
            int v=E[i].v;
            E[i].u=id[E[i].u];
            E[i].v=id[E[i].v];
            if(E[i].u!=E[i].v)
            {
                E[i].cost-=in[v];
            }
        }
        nv=cntnode;
        root=id[root];
    }
    return ret;
}

int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(int i=0;i<n;i++)
            scanf("%lf%lf",&p[i].x,&p[i].y);
        for(int i=0;i<m;i++)
        {
            scanf("%d%d",&E[i].u,&E[i].v);
            E[i].u--;
            E[i].v--;
            if(E[i].u!=E[i].v)
                E[i].cost=dis(p[E[i].u],p[E[i].v]);
            else
                E[i].cost=inf;
        }
        double ans=direct_mst(0,n,m);
        if(ans==-1)
            printf("poor snoopy\n");
        else
            printf("%.2f\n",ans);
    }
    return 0;
}