hust 1208 Playing games

题目描述

In our childhood, we all like playing games with our friends. Now you are a teacher in a kindergarten and there are n children in your class. Before playing games, you must divide the children into some small groups and each group must contains more than one child, but a child only wants to stand besides his or her friends. k (k > 1) children can form a group only if there exists a circular permutation of k children in which each child only stand besides his or her friends. To make the problem simple, given you n children and the relationship between them, you must judge if every child can be divided into some group and no one will be left alone.

输入

In the first line of the input, there is an integer T (0 < T <= 20), which means T test cases. For each test case, the first line contains two integers n (0 < n <= 400) and m (0 < m < n*n/2), which are the number of the children and the number of the relationship. There are m lines followed, every line has two integers a and b (1 <= a, b <= n ), means child a and child b are friends..

输出

For each test case, if every child can be divided into some group and no child will be left alone, output "YES", or else output "NO".

样例输入

2
4 2
1 2
3 4
3 2
1 2
1 3

样例输出

YES
NO

这个题很难想到是一个最大匹配问题，当最大匹配为n的时候就没有人独立，想谈谈二分图最大匹配的两个算法，真的效率高了很多，匈牙利算法132ms，而Hopcroft_Carp算法20ms，差别真的很大

#include<map>
#include<set>
#include<stack>
#include<queue>
#include<cmath>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define  inf 0x0f0f0f0f
 
using namespace std;
 
struct Hopcroft_Carp
{
    static const int maxn=3000+10;
    int Mx[maxn],My[maxn],Nx,Ny,dx[maxn],dy[maxn],dis;
    bool vis[maxn];
    vector<int>G[maxn];
 
    void init()
    {
        for (int i=0;i<=Nx;i++)
        G[i].clear();
    }
 
    bool BFS()
    {
        queue<int>Q;
        dis=inf;
        memset(dx,-1,sizeof(dx));
        memset(dy,-1,sizeof(dy));
        for (int i=1;i<=Nx;i++)
        if (Mx[i]==-1)
        {
            Q.push(i);
            dx[i]=0;
        }
        while(!Q.empty())
        {
            int u=Q.front(); Q.pop();
            if (dx[u]>dis) break;
            for (int i=0;i<G[u].size();i++)
            {
                int v=G[u][i];
                if (dy[v]==-1)
                {
                    dy[v]=dx[u]+1;
                    if (My[v]==-1) dis=dy[v];
                    else
                    {
                        dx[My[v]]=dy[v]+1;
                        Q.push(My[v]);
                    }
                }
            }
        }
        return dis!=inf;
    }
 
    bool DFS(int u)
    {
        for (int i=0;i<G[u].size();i++)
        {
            int v=G[u][i];
            if (!vis[v] && dy[v]==dx[u]+1)
            {
                vis[v]=true;
                if (My[v]!=-1 && dy[v]==dis) continue;
                if (My[v]==-1 || DFS(My[v]))
                {
                    My[v]=u;
                    Mx[u]=v;
                    return true;
                }
            }
        }
        return false;
    }
 
    int MaxMatch()
    {
        int ans=0;
        memset(Mx,-1,sizeof(Mx));
        memset(My,-1,sizeof(My));
        while(BFS())
        {
            memset(vis,0,sizeof(vis));
            for (int i=1;i<=Nx;i++)
            if (Mx[i]==-1 && DFS(i)) ans++;
        }
        return ans;
    }
};
 
Hopcroft_Carp friends;
 
int main()
{
    //freopen("in.txt","r",stdin);
    int t,n,m,x,y;
    scanf("%d",&t);
    while(t--)
    {
         scanf("%d%d",&n,&m);
         friends.Nx=n;friends.Ny=n;
         friends.init();
         while(m--)
         {
              scanf("%d%d",&x,&y);
              friends.G[x].push_back(y);
              friends.G[y].push_back(x);
         }
         int ans=friends.MaxMatch();
         if (ans==n) printf("YES
");
         else printf("NO
");
    }
    return 0;
}

#include<map>
#include<set>
#include<stack>
#include<queue>
#include<cmath>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define  inf 0x0f0f0f0f
 
using namespace std;
int Map[500][500];
int link[500],use[500],n,m;
bool dfs(int cap)
{
    for (int i=1;i<=m;i++)
    {
        if(Map[cap][i] && !use[i])
        {
            use[i]=1;
            int j=link[i];
            link[i]=cap;
            if(j==-1 || dfs(j)) return true;
            link[i]=j;
        }
    }
    return false;
}
int hugry()
{
    int ans=0;
    memset(link,-1,sizeof(link));
    for (int i=1;i<=n;i++)
    {
        memset(use,0,sizeof(use));
        if(dfs(i)) ans++;
    }
    return ans;
}
int main()
{
    //freopen("in.txt","r",stdin);
    int x,y,t,M;
    scanf("%d",&t);
    while(t--)
    {
         scanf("%d%d",&n,&M);
         m=n;
         memset(Map,0,sizeof(Map));
         while(M--)
         {
              scanf("%d%d",&x,&y);
              Map[x][y]=Map[y][x]=1;
         }
         int ans=hugry();
         if (ans==n) printf("YES
");
         else printf("NO
");
    }
    //fclose(stdin);
    return 0;
}

作者 chensunrise