Ant Trip

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5501    Accepted Submission(s): 2146

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

题目大意与分析

一笔画问题，求将所有的边都只走一次所需要的最少笔画数，首先用并查集维护连通图，对于每个连通图，如果所有的点度都为偶数，代表有一个欧拉路，如果只有一个点，则不需要笔画，对于有度为奇数的点，需要的笔画数为奇度点个数/2

#include<bits/stdc++.h>

using namespace std;

int f[100005],deg[100005],g[100005],j[100005],anss,cnt,n,m;

int findf(int x)
{
    return x==f[x]?x:f[x]=findf(f[x]);
}

void hebing(int a,int b)
{
    int fa=findf(a);
    int fb=findf(b);
    if(fa!=fb)
    {
        f[fa]=fb;
    }
}

int main()
{
    while(cin>>n>>m)
    {
        anss=0;
        memset(deg,0,sizeof(deg));
        memset(g,0,sizeof(g));
        memset(j,0,sizeof(j));
        for(int i=1;i<=n;i++)
        f[i]=i;
        cnt=0;
        while(m--)
        {
            int a,b;
            cin>>a>>b;
            hebing(a,b);
            deg[a]++;
            deg[b]++;
         }
        for(int i=1;i<=n;i++)
        {
            if(findf(i)==i)
            {
                g[cnt++]=i;
            }
            if(deg[i]%2==1)
            {
                j[findf(i)]++;
            }
        }
        for(int i=0;i<cnt;i++)
        {
            if(deg[g[i]]==0)
            continue;
            if(j[g[i]]==0)
            anss++;
            else
            anss+=j[g[i]]/2;
        }
        cout<<anss<<endl;
    }
 } 

原文地址：https://www.cnblogs.com/dyhaohaoxuexi/p/12635208.html