hdu3018欧拉回路题

Ant Trip

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1107    Accepted Submission(s): 404

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

 欧拉回路简单题：

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <math.h>
#include <vector>
#include <stack>
using namespace std;
#define ll long long int
int a[110000];
int aa[100005];
int find(int x)
{
    while(x!=a[x])x=a[x];
    return x;
}
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(aa,0,sizeof(aa));
        int i,x,y;
        int b[n+1];
        for(i=1; i<=n; i++)
            a[i]=i;
        memset(b,0,sizeof(b));
        for(i=0; i<m; i++)
        {
            scanf("%d%d",&x,&y);
            b[x]++;
            b[y]++;
            int fx=find(x);
            int fy=find(y);
            if(fy!=fx)
                a[fy]=fx;
        }
        int sum=0;
        for(i=1; i<=n; i++)
        {
            if(b[i]&1)
                aa[find(a[i])]++;
        }
        for(i=1; i<=n; i++)
        {
            int tt=find(a[i]);
            if(b[i]&&tt==i)
            {
                if(aa[tt])
                    sum+=aa[tt]/2;
                else sum++;
            }
        }
        cout<<sum<<endl;
    }

}