Ant Trip HDU

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

问题的关键是对于一幅图，最少需要几笔经过每一条边，并且每条边恰好只能走一次

#include<cstdio>#include<iostream>#include<algorithm>#include<cstdlib>#include<cstring>#include<cmath>#include<set>#include<vector>#define N 100005using namespace std;vector<int> graph[N];bool vis[N];int n,m;int num;void dfs(int x){    if(graph[x].size()&1)        num++;    for(int i=0;i<graph[x].size();i++)    {        int v=graph[x][i];        if(vis[v])            continue;        vis[v]=true;        dfs(v);    }}int main(){    int x,y;    while(scanf("%d%d",&n,&m)==2)    {        int ans=0;        memset(vis,false,sizeof(vis));        for(int i=1;i<=n;i++)            graph[i].clear();        while(m--)        {            scanf("%d%d",&x,&y);            graph[x].push_back(y);            graph[y].push_back(x);        }        for(int i=1;i<=n;i++)            if(!vis[i])            {                if(graph[i].size()==0)                    continue;                num=0;                vis[i]=true;                dfs(i);                if(num>0)                ans+=num/2;                else                ans+=1;                     }        printf("%d\n",ans);    }    return 0;}