HDOJ3018 欧拉回路

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#欧拉回路

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本文介绍了一个编程问题——AntTrip旅行挑战，目标是最小化蚂蚁团队的数量来遍历所有路径。文章探讨了如何通过并查集算法确定连通分量，并根据不同连通分量的特点计算所需的最小蚂蚁团队数。

Ant Trip

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3317 Accepted Submission(s): 1324

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

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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假如连通分量是个孤立的点，需要零笔。

假如连通分量是个（半）欧拉图，需要一笔。

假如连通分量是个非（半）欧拉图，需要笔数为该图中奇数度点的数目/2

用一个num数组统计每个连通分量的个数。

假如num【i】==0，表示该节点不是根节点。

假如num【i】==1 表示该节点孤立存在。

假如num【i】>1，再分是否为（半）欧拉图计算结果。

#include <iostream>
#include <cstdio>
using namespace std;

const int maxn = 1e5+5;
int du[maxn],f[maxn],num[maxn],odd[maxn];
int i,j,k,n,m,cnt,a,b;

int Find(int x){
    if (f[x]==x) return x;
    return f[x] = Find(f[x]);
}

void Merge(int x, int y){
   int fx = Find(x), fy = Find(y);
   if (fx!=fy){
      f[fx]=fy;
   }
}

int main(){
    while (~scanf("%d%d",&n,&m)){
         for (i=1; i<=n; i++) {
            du[i] = 0;
            odd[i] = 0;
            f[i] = i;
            num[i] = 0;
         }

         for (i=1; i<=m; i++) {
            scanf("%d%d",&a,&b);
            du[a]++;
            du[b]++;
            Merge(a,b);
         }

         for (i=1; i<=n; i++) num[Find(i)]++;
         for (i=1; i<=n; i++) if (du[i]&1) odd[Find(i)]++;


         cnt = 0;
         for (i=1; i<=n; i++) if (num[i]>1){
                if (odd[i]>0) cnt+=odd[i]/2;
                else cnt++;
         }
         printf("%d\n",cnt);
    }
    return 0;
}