poj 3164 Command Network（最小树形图朱刘算法）

题目链接：http://poj.org/problem?id=3164

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 integerN (N ≤ 100), the number of nodes in the destroyed network, andM (M ≤ 10⁴), the number of pairs of nodes between which a wire can be built. The nextN lines each contain an ordered pairx_i and y_i, giving the Cartesian coordinates of the nodes. Then followM lines each containing two integersi and j between 1 andN (inclusive) meaning a wire can be built between nodei and nodej 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

Source

POJ Monthly--2006.12.31, galaxy

题意，是给定你n个点的坐标，然后是m个这些点的关系，用点的距离作为花费，建图，然后直接求最小树形图，经典的朱刘算法求解。

代码：

#include<stdio.h>#include<string.h>#include<math.h>const int maxn=1100;const int inf=1<<31-1;struct  point{    int x,y;}P[maxn*maxn];struct node{    int u,v;    double cost;}E[maxn*maxn];int pre[maxn],vis[maxn],ID[maxn];double In[maxn];int n,m;double get_dis(int s,int t){    int x1=P[s].x;    int y1=P[s].y;    int x2=P[t].x;    int y2=P[t].y;    return sqrt((x1-x2)*(x1-x2)*1.0+(y1-y2)*(y1-y2)*1.0);}double zhuliu(int root,int nv,int ne){    double ret=0;    while(true)    {        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;        }        int cnt=0;        memset(ID,-1,sizeof(ID));        memset(vis,-1,sizeof(vis));        In[root]=0;        for(int i=0;i<nv;i++)        {            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]=cnt;                ID[v]=cnt++;            }        }        if(cnt==0) break;        for(int i=0;i<nv;i++)        {            if(ID[i]==-1)                ID[i]=cnt++;        }        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=cnt;        root=ID[root];    }    return ret;}int main(){    //freopen("in.txt","r",stdin);    while(scanf("%d %d",&n,&m)==2)    {        for(int i=0;i<n;i++)            scanf("%d %d",&P[i].x,&P[i].y);        for(int i=0;i<m;i++)        {            int s,t;            scanf("%d %d",&s,&t);            E[i].u=s-1;            E[i].v=t-1;            E[i].cost=get_dis(s-1,t-1);            //printf("cost=%lf ",E[i].cost);        }//        for(int i=0;i<m;i++)//            printf("%d %d %lf\n",E[i].u,E[i].v,E[i].cost);        double ans=zhuliu(0,n,m);        if(ans==-1)            printf("poor snoopy\n");        else            printf("%.2lf\n",ans);    }    return 0;}