zoukankan      html  css  js  c++  java
  • POJ 3164 Command Network(最小树形图)

    Command Network
    Time Limit: 1000MS   Memory Limit: 131072K
    Total Submissions: 13817   Accepted: 3973

    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 xi and yi, 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

    裸最小树形图。

    #include <iostream>
    #include <algorithm>
    #include <cstdio>
    #include <cmath>
    #include <cstring>
    #include <vector>
    
    using namespace std ;
    typedef long long LL ;
    const double inf = 1e9+7;
    const int N = 1010;
    const int M = 40010;
    
    struct edge {
        int u , v ;
        double w ;
    }e[M];
    int n , m ;
    double x[N] , y[N] ;
    int pre[N] , id[N] , vis[N] ;
    double in[N] ;
    
    double zhuliu( int root ) {
        double res = 0 ; int u , v ;
        while(1) {
            for( int i = 0 ; i < n ; ++i ) in[i] = inf ;
    
            for( int i = 0 ; i < m ; ++i )
                if( e[i].u != e[i].v && e[i].w < in[e[i].v] ) {
                    pre[e[i].v] = e[i].u;
                    in[e[i].v] = e[i].w;
                }
            for( int i = 0 ; i < n ; ++i )
                if( i != root && in[i] == inf )
                    return -1 ;
            int tn = 0 ;
            memset( id , -1 , sizeof id );
            memset( vis , -1 , sizeof vis );
            in[root] = 0 ;
            for( int i = 0 ; i < n ; ++i ) {
                res += in[i];
                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] = tn ;
                    id[v] = tn++ ;
                }
            }
            if( tn == 0 ) break ; // no circle
            for( int i = 0 ; i < n ; ++i ) if( id[i] == -1 ) {
                id[i] = tn++ ;
            }
            for( int i = 0 ; i < m ; ){
                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++].w -= in[v];
                else
                    swap( e[i] , e[--m] );
            }
            n = tn ;
            root = id[root];
        }
        return res ;
    }
    int main ()  {
    //    freopen("in.txt","r",stdin);
        while( ~scanf("%d%d",&n,&m) ) {
            for( int i = 0 ; i < n ; ++i )
                scanf("%lf%lf",&x[i],&y[i]);
            for( int i = 0 ; i < m ; ++i ) {
                int u , v ; scanf("%d%d",&u,&v);
                u -- , v -- ;
                e[i].u = u ; e[i].v = v ;
                e[i].w = sqrt((x[u]-x[v])*(x[u]-x[v])+(y[u]-y[v])*(y[u]-y[v]));
            }
            double ans = zhuliu( 0 );
            if( ans == -1 )
                puts("poor snoopy");
            else
                printf("%.2lf
    ",ans);
        }
    }
    View Code
    only strive for your goal , can you make your dream come true ?
  • 相关阅读:
    Linux_23 DNS 正向解析区域、反向解析区域;主/从;子域;基本安全控制
    Linux_22 加密和解密及OpenSSL
    Linux_21 日志系统、ssh服务、系统安装及系统故障排除
    Linux_20 子网划分
    Akavache简明使用指南
    Oracle存储过程解析XML内容
    docker部署微服务不支持中文字体的解决方案
    Three.js
    Three.js
    [Linux] vim状态栏配置
  • 原文地址:https://www.cnblogs.com/hlmark/p/4292081.html
Copyright © 2011-2022 走看看