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

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

2009 Multi-University Training Contest 12 – Host by FZU

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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\n",ans);

   }
    return 0;
}

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