zoukankan      html  css  js  c++  java
  • Road Construction 求双连通分量(tarjan)缩点,最后求度为1的点的个数cnt,则所需要增加的边(cnt+1)/2

    Problem Description

    It's almost summer time, and that means that it's almost summer construction time! This year, the good people who are in charge of the roads on the tropical island paradise of Remote Island would like to repair and upgrade the various roads that lead between the various tourist attractions on the island.

    The roads themselves are also rather interesting. Due to the strange customs of the island, the roads are arranged so that they never meet at intersections, but rather pass over or under each other using bridges and tunnels. In this way, each road runs between two specific tourist attractions, so that the tourists do not become irreparably lost.

    Unfortunately, given the nature of the repairs and upgrades needed on each road, when the construction company works on a particular road, it is unusable in either direction. This could cause a problem if it becomes impossible to travel between two tourist attractions, even if the construction company works on only one road at any particular time.

    So, the Road Department of Remote Island has decided to call upon your consulting services to help remedy this problem. It has been decided that new roads will have to be built between the various attractions in such a way that in the final configuration, if any one road is undergoing construction, it would still be possible to travel between any two tourist attractions using the remaining roads. Your task is to find the minimum number of new roads necessary.

     

    Input

    The first line of input will consist of positive integers n and r, separated by a space, where 3 ≤ n ≤ 1000 is the number of tourist attractions on the island, and 2 ≤ r≤ 1000 is the number of roads. The tourist attractions are conveniently labelled from 1 to n. Each of the following r lines will consist of two integers, v and w, separated by a space, indicating that a road exists between the attractions labelled v and w. Note that you may travel in either direction down each road, and any pair of tourist attractions will have at most one road directly between them. Also, you are assured that in the current configuration, it is possible to travel between any two tourist attractions.

     

    Output

    One line, consisting of an integer, which gives the minimum number of roads that we need to add.

     

    Sample Input
    Sample Input 1 10 12 1 2 1 3 1 4 2 5 2 6 5 6 3 7 3 8 7 8 4 9 4 10 9 10 Sample Input 2 3 3 1 2 2 3 1 3
     

    Sample Output
    Output for Sample Input 1 2 Output for Sample Input 2 0
     **************************************************************************************************************************
    缩点的实质就是标记同一连通块
    ***************************************************************************************************************************
    #include<iostream>
    #include<string>
    #include<cstring>
    #include<queue>
    #include<cstdio>
    #include<vector>
    using namespace std;
    vector<int>adj[2001];
    int vis[2001],low[2001],degree[2001];
    bool  co[2001][2001];
    int n,k,i,j,ed,cnt;
    void init()
    {
        memset(vis,0,sizeof(vis));
        memset(low,0,sizeof(low));
        memset(degree,0,sizeof(degree));
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=n;j++)
            {
                co[i][j]=false;
            }
            adj[i].clear();
        }
    
    }
    void dfs(int x,int fa)
    {
        vis[x]=true;
        low[x]=cnt++;
        int siz=adj[x].size();
        for(int it=0;it<siz;it++)
        {
            int cur=adj[x][it];
            if(adj[x][it]==fa)
              continue;
            if(!vis[cur])
            {
                dfs(cur,x);
            }
            if(low[x]>low[cur])
              low[x]=low[cur];
        }
    }
    int solve()
    {
        int cnt=0;
        for(i=1;i<=n;i++)
        {
            int lent=adj[i].size();
            for(j=0;j<lent;j++)
              if(low[adj[i][j]]!=low[i])
                degree[low[i]]++;
        }
        for(i=1;i<=n;i++)
          if(degree[i]==1)
            cnt++;
        return (cnt+1)/2;
    }
    int main()
    {
        while(scanf("%d%d",&n,&ed)!=EOF)
        {
            int st,en;
            init();
            for(i=1;i<=ed;i++)
            {
                scanf("%d%d",&st,&en);
                if(!co[st][en])
                {
                    adj[st].push_back(en);
                    adj[en].push_back(st);
                    co[st][en]=true;
                }
            }
            dfs(1,1);
            printf("%d
    ",solve());
        }
        return 0;
    }
  • 相关阅读:
    nginx重启失败的解决方法
    Linux开机禁用开启防火墙
    Linux查看系统版本信息的几种方法
    Linux修改时区的正确方法
    Linux中awk的gsub函数用法
    Linux 的mktemp 命令详解
    tomcat的简单概要小结
    {面试题8: 旋转数组的最小数字}
    {面试题7: 使用两个队列实现一个栈}
    {面试题7: 用两个栈实现队列}
  • 原文地址:https://www.cnblogs.com/sdau--codeants/p/3365755.html
Copyright © 2011-2022 走看看