hdu 3018 Ant Trip 欧拉回路+并查集

Ant Trip

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

题意：给你n个城市，m条边，一个人只能走他没有走过的路，问最少几个人才能把路走完；

思路：相当于只需要求有多少条欧拉路径，利用并查集判联通，如果一个图联通的并且含有n个奇数度的点，那么需要走n/2,因为一条欧拉路径最多两个奇数点;

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define mod 1000000007
#define esp 0.00000000001
const int N=1e5+10,M=1e6+10,inf=1e9;
const ll INF=1e18+10;
int n,m;
int fa[N],du[N],flag[N],mark[N],hh[N];
int Find(int x)
{
    return fa[x]==x?x:fa[x]=Find(fa[x]);
}
void update(int u,int v)
{
    int x=Find(u);
    int y=Find(v);
    if(x!=y)
    {
        fa[x]=y;
    }
}
void init()
{
    for(int i=0;i<N;i++)
        fa[i]=i;
    memset(flag,0,sizeof(flag));
    memset(du,0,sizeof(du));
    memset(mark,0,sizeof(mark));
    memset(hh,0,sizeof(hh));
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        init();
        for(int i=1;i<=m;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            update(u,v);
            du[u]++;
            du[v]++;
            hh[u]=1;
            hh[v]=1;
        }
        int ans=0,sum=0;
        for(int i=1;i<=n;i++)
        {
            if(!hh[i])continue;
            int x=Find(i);
            du[i]%=2;
            sum+=du[i];
            if(!flag[x])
            {
                ans++;
                flag[x]=1;
            }
            if(du[i]==1&&flag[x]&&!mark[x])
            {
                ans--;
                mark[x]=1;
            }
        }
        printf("%d
",ans+sum/2);
    }
    return 0;
}