Ant Trip

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2010    Accepted Submission(s): 782

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

2009 Multi-University Training Contest 12 - Host by FZU

一个连通图不会有奇数个奇数度顶点，画一画图可以知道，假如一个连通图所有点度数为偶数，答案为1，否则就是奇数度顶点的一半，累加求和。

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
const  int maxn=100005;
int father[maxn];
int indegree[maxn];
int n,m;
int a[maxn];
void init(){
    for(int i=1;i<=n;i++)
        father[i]=i;
}
int find(int u){
   if(father[u]!=u)
    father[u]=find(father[u]);
   return father[u];
}
void Union(int x,int y){
     int t1=find(x);
     int t2=find(y);
     if(t1!=t2)
        father[t1]=t2;
}
int main(){
   while(scanf("%d%d",&n,&m)!=EOF){
        init();
      // memset(father,0,sizeof(father));
       memset(indegree,0,sizeof(indegree));
       memset(a,-1,sizeof(a));
       for(int i=1;i<=m;i++){
           int u,v;
          scanf("%d%d",&u,&v);
          indegree[u]++;
          indegree[v]++;
           Union(u,v);
       }

       for(int i=1;i<=n;i++){
           int t=find(i);
           if(a[t]==-1&&indegree[i]>0)
            a[t]=0;
           if(indegree[i]&1)
            a[t]++;
       }
int ans=0;
       for(int i=1;i<=n;i++){
        if(a[i]==0)
            ans++;
        else if(a[i]>0)
            ans+=a[i]/2;
       }
       printf("%d\n",ans);
   }
   return 0;
}