HDU 3018 Ant Trip（欧拉路径 + 并查集 + 连通性判断）

最新推荐文章于 2019-03-18 20:48:45 发布

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最新推荐文章于 2019-03-18 20:48:45 发布

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分类专栏：图的可行遍性并查集

本文链接：https://blog.youkuaiyun.com/qwe585p/article/details/48182955

并查集同时被 2 个专栏收录

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图的可行遍性

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本文探讨了如何通过解决一笔画问题，即在一个图中，仅使用最少的笔触来遍历所有边，进而解决在AntCountry中，Tony和他的朋友们想要遍历整个国家，每条道路只访问一次的问题。文章详细解释了求连通图和欧拉图的方法，并通过实例展示了计算最少需要几组蚂蚁来完成这一任务。

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题意：问给定一张图，每次蚂蚁能经过一条未经过的边，问经过整张图，需要几次。其实就是一笔画问题，问最少需要几笔

解题思路：脑卡了一下，看看题解会了。

就是求连通图和欧拉图的个数， ans = 欧拉图个数 + 连通图奇数度点的个数 / 2；其中连通图包含了欧拉路径

不错的题目。

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.

输入

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.

输出

For each test case ,output the least groups that needs to form to achieve their goal.

样例输入

#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<cctype>
#include<list>
#include<iostream>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>

using namespace std;

#define FOR(i, s, t) for(int i = (s) ; i <= (t) ; ++i)
#define REP(i, n) for(int i = 0 ; i < (n) ; ++i)

int buf[10];
inline long long read()
{
    long long x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9')
    {
        if(ch=='-')f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}

inline void writenum(int i)
{
    int p = 0;
    if(i == 0) p++;
    else while(i)
        {
            buf[p++] = i % 10;
            i /= 10;
        }
    for(int j = p - 1 ; j >= 0 ; --j) putchar('0' + buf[j]);
}
/**************************************************************/
#define MAX_N 100005
const int INF = 0x3f3f3f3f;
int parent[MAX_N];
int degree[MAX_N];
int vis[MAX_N];
inline int init(int n)
{
    memset(vis, -1, sizeof(vis));
    memset(degree, 0, sizeof(degree));
    for(int i = 1 ; i < n + 5 ; i++)
    {
        parent[i] = i;
    }
}

int find(int a)
{
    if(parent[a] != a)
    {
        parent[a] = find(parent[a]);
    }
    return parent[a];
}

void unite(int a, int b)
{
    int fa = find(a);
    int fb = find(b);
    if(fa != fb)
    {
        parent[fb] = fa;
    }
    return;
}

int main()
{
    int n, m;
    while(~scanf("%d%d", &n ,&m))
    {
        int ans = 0;
        init(n);
        int u, v;
        for(int i = 0 ; i < m ; i++)
        {
            u = read();
            v = read();
            unite(u, v);
            degree[u]++;
            degree[v]++;
        }
        for(int i = 1 ; i <= n ; i++)
        {
            int tmp = find(i);
//            cout<<"tmp"<<tmp<<endl;
            if(vis[tmp] == -1 && degree[i])
            {
                vis[tmp] = 0;
            }
            if(degree[i] % 2)
            {
                vis[tmp]++;
            }
//            cout<<"vis[tmp]"<<vis[tmp]<<endl;
        }
        for(int i = 1 ; i <= n ; i++)
        {
//            cout<<vis[i]<<endl;
            if(vis[i] > 0) ans += vis[i] / 2;
            else if(!vis[i]) ans++;
        }
        printf("%d\n", ans);
    }
    return 0;
}