HDu 3018 Ant Trip(オイラーバック)

Ant Trip
Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 14   Accepted Submission(s) : 8
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. 
   

  

 
          実はテーマの意味はあなたに1枚の図をあげて、少なくともどれだけのペンで全体の図を描くことができて、すべての辺は1回しか通ることができなくて、孤立点は無視します.これがオラ回路の問題で、何筆かを漕ぐには入度と出度の値、そして入度-出度の値を見て、どれだけのオラ路があるかを判断します.連通性を並列調査集で判断し,度数でオーラループを判断する.あまり言わないで、次のコードです.
リンク:http://acm.hdu.edu.cn/showproblem.php?pid=3018
コード:

#include <iostream>
#include <stdio.h>
#include <memory.h>
#include <vector>
using namespace std;

int father[100005], deg[100005], odd[100005];
bool hash[100005];
vector<int> a;
int n, m, sum;

void init()
{
    int i;
    a.clear();
    memset(hash, false, sizeof(hash));
    memset(deg, 0, sizeof(deg));
    memset(odd, 0, sizeof(odd));
    for(i = 1; i <= n; i++) father[i] = i;
}

int find(int x)
{
    if(x != father[x])
    {
        father[x] = find(father[x]);
    }
    return father[x];
}

void Union(int x, int y)
{
    father[x] = y;
}

int main()
{
    int i, k, x, y, fx, fy;
    while(scanf("%d %d", &n, &m) != EOF)
    {
        init();
        for(i = 1; i <= m; i++)
        {
            scanf("%d %d", &x, &y);
            deg[x]++;
            deg[y]++;
            fx = find(x);
            fy = find(y);
            if(fx != fy) Union(fx, fy);
        }
        for(i = 1; i <= n; i++)
        {
            k = find(i);
            if(!hash[k])
            {
                a.push_back(k); //a     ，     
                hash[k] = true;
            }
            if(deg[i]%2 == 1) odd[k]++; //              
        }
        sum = 0;
        for(i = 0; i < a.size(); i++)
        {
            k = a[i];
            if(deg[k] == 0) continue;   //   
            if(odd[k] == 0) sum++;      //        ，    
            else sum += odd[k]/2;
        }
        printf("%d
", sum);
    }

    return 0;
}