HDU 3108 Ant Trip

Ant Trip

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3524    Accepted Submission(s):
1393

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

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题意：给出N个节点，M个边，问要遍历一遍所有的边，需要的最小group数目。

思路：首先通过并查集判断有几个联通块。然后对于每个联通块，统计需要多少笔。

如果，联通块中，度为奇数节点的个数为0或2个，那就可以一笔画完。

反之，因为一笔最多消去两个度为奇数的节点，所以要最少用奇度数节点个数/2笔。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int n,m,tot,ans;
int vis[];
int fa[],num[];
int into[],du[];
int find(int x){
    if(fa[x]==x)    return fa[x];
    else return fa[x]=find(fa[x]);
}
int main(){
    while(scanf("%d%d",&n,&m)!=EOF){
        tot=;ans=;
        memset(du,,sizeof(du));
        memset(vis,,sizeof(vis));
        memset(num,,sizeof(num));
        memset(into,,sizeof(vis));
        for(int i=;i<=n;i++)    fa[i]=i;
        for(int i=;i<=m;i++){
            int x,y;
            scanf("%d%d",&x,&y);
            into[x]++;into[y]++;
            int dx=find(x);int dy=find(y);
            if(dx!=dy)    fa[dy]=dx;
        }
        for(int i=;i<=n;i++){
            int now=find(i);
            if(!vis[now]){
                vis[now]=;
                num[++tot]=now;
            }
            if(into[i]%!=)    du[now]++;
        }
        for(int i=;i<=tot;i++){
            if(into[num[i]]==)    continue;
            if(du[num[i]]==)    ans++;
            ans+=du[num[i]]/;
        }
        cout<<ans<<endl;
    }
}