zoukankan      html  css  js  c++  java
  • POJ1523 SPF[无向图割点]

    SPF
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 8139   Accepted: 3723

    Description

    Consider the two networks shown below. Assuming that data moves around these networks only between directly connected nodes on a peer-to-peer basis, a failure of a single node, 3, in the network on the left would prevent some of the still available nodes from communicating with each other. Nodes 1 and 2 could still communicate with each other as could nodes 4 and 5, but communication between any other pairs of nodes would no longer be possible. 

    Node 3 is therefore a Single Point of Failure (SPF) for this network. Strictly, an SPF will be defined as any node that, if unavailable, would prevent at least one pair of available nodes from being able to communicate on what was previously a fully connected network. Note that the network on the right has no such node; there is no SPF in the network. At least two machines must fail before there are any pairs of available nodes which cannot communicate. 

    Input

    The input will contain the description of several networks. A network description will consist of pairs of integers, one pair per line, that identify connected nodes. Ordering of the pairs is irrelevant; 1 2 and 2 1 specify the same connection. All node numbers will range from 1 to 1000. A line containing a single zero ends the list of connected nodes. An empty network description flags the end of the input. Blank lines in the input file should be ignored.

    Output

    For each network in the input, you will output its number in the file, followed by a list of any SPF nodes that exist. 

    The first network in the file should be identified as "Network #1", the second as "Network #2", etc. For each SPF node, output a line, formatted as shown in the examples below, that identifies the node and the number of fully connected subnets that remain when that node fails. If the network has no SPF nodes, simply output the text "No SPF nodes" instead of a list of SPF nodes. 

    Sample Input

    1 2
    5 4
    3 1
    3 2
    3 4
    3 5
    0
    
    1 2
    2 3
    3 4
    4 5
    5 1
    0
    
    1 2
    2 3
    3 4
    4 6
    6 3
    2 5
    5 1
    0
    
    0

    Sample Output

    Network #1
      SPF node 3 leaves 2 subnets
    
    Network #2
      No SPF nodes
    
    Network #3
      SPF node 2 leaves 2 subnets
      SPF node 3 leaves 2 subnets

    Source


    cut vertex裸题,只不过求的是能分成几个subnets,使成为割点的子节点数+1就行(还有父辈们一个),注意root不能加1
    输入输出格式有点坑
    //
    //  main.cpp
    //  poj1523
    //
    //  Created by Candy on 9/26/16.
    //  Copyright © 2016 Candy. All rights reserved.
    //
    
    #include <iostream>
    #include <cstdio>
    #include <algorithm>
    #include <cstring>
    using namespace std;
    const int N=1e3+5,INF=1e9+5;
    inline int read(){
        char c=getchar();int x=0,f=1;
        while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
        while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
        return x;
    }
    int n=0,u,v,w,cas=0;
    struct edge{
        int v,ne;
    }e[N*N];
    int h[N],cnt=0;
    inline void ins(int u,int v){
        cnt++;
        e[cnt].v=v;e[cnt].ne=h[u];h[u]=cnt;
        cnt++;
        e[cnt].v=u;e[cnt].ne=h[v];h[v]=cnt;
    }
    int pre[N],sn[N],dc=0;
    int dfs(int u,int fa){
        int lowu=pre[u]=++dc,child=0;
        for(int i=h[u];i;i=e[i].ne){
            int v=e[i].v;
            if(!pre[v]){
                child++;
                int lowv=dfs(v,u);
                lowu=min(lowu,lowv);
                if(lowv>=pre[u]) sn[u]++;//,printf("v %d %d
    ",u,v);
            }else if(pre[v]<pre[u]&&v!=fa)
                lowu=min(lowu,pre[v]);
        }
        if(fa==-1&&child==1) sn[u]=0;
        return lowu;
    }
    int main(int argc, const char * argv[]) {
        while((u=read())){
            int root=0,flag=0; n=0;
            memset(h,0,sizeof(h)); cnt=0;
            memset(pre,0,sizeof(pre));
            memset(sn,0,sizeof(sn));
            v=read();
            ins(u,v);
            n=max(n,max(u,v));
            while((u=read())){
                v=read();
                ins(u,v);
                n=max(n,max(u,v)); root=u;
            }
            dfs(root,-1);
            if(sn[root]!=0) sn[root]--;
            printf("Network #%d
    ",++cas);
            for(int i=0;i<=n;i++) if(sn[i]){
                printf("  SPF node %d leaves %d subnets
    ",i,sn[i]+1);
                flag=1;
            }
            if(!flag) printf("  No SPF nodes
    ");
            printf("
    ");
        }
        
        return 0;
    }
  • 相关阅读:
    光线步进——RayMarching入门
    MATLAB GUI制作快速入门
    Python中用绘图库绘制一条蟒蛇
    node 常见的一些系统问题
    webpack 入门指南
    利用 gulp 来合并seajs 的项目
    移动端 解决自适应 和 多种dpr (device pixel ratio) 的 [淘宝] 解决方案 lib-flexible
    富有哲理的文章
    NodeJS 难点(网络,文件)的 核心 stream 四: writable
    Vue.js 源码学习笔记 -- 分析前准备2 -- Object.defineProperty
  • 原文地址:https://www.cnblogs.com/candy99/p/5911237.html
Copyright © 2011-2022 走看看