zoukankan      html  css  js  c++  java
  • [欧拉回路] hdu 3018 Ant Trip

    题目链接:

    http://acm.hdu.edu.cn/showproblem.php?pid=3018

    Ant Trip

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 1658    Accepted Submission(s): 641


    Problem Description
    Ant Country consist of N towns.There are M roads connecting the towns.

    Ant Tony,together with his friends,wants to go through every part of the country. 

    They intend to visit every road , and every road must be visited for exact one time.However,it may be a mission impossible for only one group of people.So they are trying to divide all the people into several groups,and each may start at different town.Now tony wants to know what is the least groups of ants that needs to form to achieve their goal.
     

    Input
    Input contains multiple cases.Test cases are separated by several blank lines. Each test case starts with two integer N(1<=N<=100000),M(0<=M<=200000),indicating that there are N towns and M roads in Ant Country.Followed by M lines,each line contains two integers a,b,(1<=a,b<=N) indicating that there is a road connecting town a and town b.No two roads will be the same,and there is no road connecting the same town.
     

    Output
    For each test case ,output the least groups that needs to form to achieve their goal.
     

    Sample Input
    3 3 1 2 2 3 1 3 4 2 1 2 3 4
     

    Sample Output
    1 2
    Hint
    New ~~~ Notice: if there are no road connecting one town ,tony may forget about the town. In sample 1,tony and his friends just form one group,they can start at either town 1,2,or 3. In sample 2,tony and his friends must form two group.
     

    Source
     

    Recommend
    gaojie   |   We have carefully selected several similar problems for you:  3013 3015 3016 3011 3010 
     

    Statistic | Submit | Discuss | Note

    题目意思:

    给一幅无向图,求要用多少次一笔画,把全部边走完,边仅仅能走一次。孤立点不算。

    解题思路:
    dfs把每一个连通块找到,然后统计奇数度数节点个数。

    注意孤立节点不算。

    代码:

    //#include<CSpreadSheet.h>
    
    #include<iostream>
    #include<cmath>
    #include<cstdio>
    #include<sstream>
    #include<cstdlib>
    #include<string>
    #include<string.h>
    #include<cstring>
    #include<algorithm>
    #include<vector>
    #include<map>
    #include<set>
    #include<stack>
    #include<list>
    #include<queue>
    #include<ctime>
    #include<bitset>
    #include<cmath>
    #define eps 1e-6
    #define INF 0x3f3f3f3f
    #define PI acos(-1.0)
    #define ll __int64
    #define LL long long
    #define lson l,m,(rt<<1)
    #define rson m+1,r,(rt<<1)|1
    #define M 1000000007
    //#pragma comment(linker, "/STACK:1024000000,1024000000")
    using namespace std;
    
    #define Maxn 110000
    int de[Maxn],n,m;
    vector<vector<int> >myv;
    int in[Maxn],cnt;
    bool vis[Maxn];
    
    void dfs(int cur)
    {
        in[++cnt]=cur;
        vis[cur]=true;
        for(int i=0;i<myv[cur].size();i++)
        {
            int ne=myv[cur][i];
            if(vis[ne])
                continue;
            dfs(ne);
        }
    }
    int main()
    {
        //freopen("in.txt","r",stdin);
       //freopen("out.txt","w",stdout);
       while(~scanf("%d%d",&n,&m))
       {
           myv.clear();
           myv.resize(n+10);
           memset(de,0,sizeof(de));
    
           for(int i=1;i<=m;i++)
           {
               int a,b;
               scanf("%d%d",&a,&b);
               myv[a].push_back(b);
               myv[b].push_back(a);
               de[a]++;
               de[b]++;
           }
           memset(vis,false,sizeof(vis));
    
           int ans=0;
    
           for(int i=1;i<=n;i++)
           {
               if(!vis[i])
               {
                   cnt=0;
                   dfs(i);
                   int temp=0;
                   if(cnt==1)  //孤立节点不算
                        continue;
                   for(int j=1;j<=cnt;j++)
                   {
                       if(de[in[j]]&1)
                            temp++;
                        //printf("i:%d j")
                   }
                   if(!temp)
                        ans++;
                   else
                        ans+=temp/2;
               }
           }
           printf("%d
    ",ans);
    
       }
        return 0;
    }
    
    


  • 相关阅读:
    快速入门:BUMO 节点安装运维指南
    快速入门:BUMO 智能合约(hello world)
    python 多进程处理 multiprocessing模块
    python 多进程处理 multiprocessing模块
    python 多进程处理 multiprocessing模块
    一段简单的数据加密小例程
    一段简单的数据加密小例程
    一段简单的数据加密小例程
    一段简单的数据加密小例程
    家用PC机打造VSphere5.1 测试环境:之部署VCenter Server 5.1
  • 原文地址:https://www.cnblogs.com/yxwkf/p/4550313.html
Copyright © 2011-2022 走看看