POJ3164-Command Network

Command Network

Time Limit: 1000MS Memory Limit: 131072KTotal Submissions: 17829 Accepted: 5111

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 iand 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

Source

POJ Monthly--2006.12.31, galaxy

题意：有N个点，M条有向边。以1为根构造出一棵最小生成树，问这棵最小生成树能否被构造出来，如果可以，总权值是多少
解题思路：用的是double型的，输出时是%.2lf，结果是WA ，换成了%.2f就A了

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <algorithm>#include <cmath>#include <queue>#include <vector>#include <set>#include <stack>#include <map>#include <climits>using namespace std;#define LL long longconst int INF=0x3f3f3f3f;const int MAXN = 1010;const int MAXM = 40010;struct Edge{    int u,v;    double cost;}edge[MAXM];struct Point{    double x,y;}p[105];int pre[MAXN],id[MAXN],visit[MAXN];double in[MAXN];double zhuliu(int root,int n,int m,Edge edge[]){    double res = 0;    int u,v;    while(1)    {        for(int i = 0; i < n; i++) in[i] = 1.0*INF;        for(int i = 0; i < m; i++)        {            if(edge[i].u != edge[i].v && edge[i].cost < in[edge[i].v])            {                pre[edge[i].v] = edge[i].u;                in[edge[i].v] = edge[i].cost;            }        }        for(int i = 0; i < n; i++)            if(i != root && in[i] == 1.0*INF) return -1; //不存在最小树形图        int tn = 0;        memset(id,-1,sizeof(id));        memset(visit,-1,sizeof(visit));        in[root] = 0;        for(int i = 0; i < n; i++)        {            res += in[i];            v = i;            while( visit[v] != i && id[v] == -1 && v != root)            {                visit[v] = i;                v = pre[v];            }            if( v != root && id[v] == -1 )            {                for(u = pre[v]; u != v ; u = pre[u])                    id[u] = tn;                id[v] = tn++;            }        }        if(tn == 0) break;//没有有向环        for(int i = 0; i < n; i++)            if(id[i]==-1) id[i]=tn++;        for(int i = 0; i < m;i++)        {            v = edge[i].v;            edge[i].u = id[edge[i].u];            edge[i].v = id[edge[i].v];            if(edge[i].u != edge[i].v) edge[i].cost -= in[v];            //else swap(edge[i],edge[--m]);        }        n = tn;        root = id[root];    }    return res;}int main(){    int n,m;    while(~scanf("%d%d",&n,&m))    {        for(int i=0;i<n;i++)            scanf("%lf %lf",&p[i].x,&p[i].y);        int u,v;        int L = 0;        while(m--)        {            scanf("%d%d",&u,&v);            u--,v--;            double dis=1.0*sqrt((p[u].x-p[v].x)*(p[u].x-p[v].x)+(p[u].y-p[v].y)*(p[u].y-p[v].y));            edge[L].u = u;            edge[L].v = v;            edge[L++].cost =dis;        }        double ans=zhuliu(0,n,L,edge);        if(ans==-1) printf("poor snoopy\n");        else printf("%.2lf\n",ans);    }    return 0;}