码迷,mamicode.com
首页 > 其他好文 > 详细

The Accomodation of Students HDU - 2444(判断二分图 + 二分匹配)

时间:2018-07-14 16:24:19      阅读:128      评论:0      收藏:0      [点我收藏+]

标签:vector   style   using   main   output   problem   src   pre   close   

The Accomodation of Students

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8418    Accepted Submission(s): 3709


Problem Description
There are a group of students. Some of them may know each other, while others don‘t. For example, A and B know each other, B and C know each other. But this may not imply that A and C know each other.

Now you are given all pairs of students who know each other. Your task is to divide the students into two groups so that any two students in the same group don‘t know each other.If this goal can be achieved, then arrange them into double rooms. Remember, only paris appearing in the previous given set can live in the same room, which means only known students can live in the same room.

Calculate the maximum number of pairs that can be arranged into these double rooms.
 

 

Input
For each data set:
The first line gives two integers, n and m(1<n<=200), indicating there are n students and m pairs of students who know each other. The next m lines give such pairs.

Proceed to the end of file.

 

 

Output
If these students cannot be divided into two groups, print "No". Otherwise, print the maximum number of pairs that can be arranged in those rooms.
 

 

Sample Input
4 4 1 2 1 3 1 4 2 3 6 5 1 2 1 3 1 4 2 5 3 6
 

 

Sample Output
No 3
 

 

Source
 

 

Recommend
gaojie

 

 

 

先判断能不能分成二分图 , 不是就输出No。。只有1个的时候也不是

然后求完美匹配即可。。。我不会写匈牙利了。。。。只会写hk。。。还是套模板。。。

因为没有分左右  所以要除2 

 

技术分享图片
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <queue>
#include <vector>
#define mem(a, b) memset(a, b, sizeof(a))
using namespace std;
const int maxn = 10010, INF = 0x7fffffff;
int dx[maxn], dy[maxn], cx[maxn], cy[maxn], used[maxn], vis[maxn];
int nx, ny, dis;
vector<int> G[40005];
int n, m;
int bfs()
{
    queue<int> Q;
    dis = INF;
    mem(dx, -1);
    mem(dy, -1);
    for(int i=1; i<=nx; i++)
    {
        if(cx[i] == -1)
        {
            Q.push(i);
            dx[i] = 0;
        }
    }
    while(!Q.empty())
    {
        int u = Q.front(); Q.pop();
        if(dx[u] > dis) break;
        for(int v=0; v<G[u].size(); v++)
        {
            int i = G[u][v];
            if(dy[i] == -1)
            {
                dy[i] = dx[u] + 1;
                if(cy[i] == -1) dis = cy[i];
                else
                {
                    dx[cy[i]] = dy[i] + 1;
                    Q.push(cy[i]);
                }
            }
        }
    }
    return dis != INF;
}

int dfs(int u)
{
    for(int v=0; v<G[u].size(); v++)
    {
        int i=G[u][v];
        if(!used[i] && dy[i] == dx[u] + 1)
        {
            used[i] = 1;
            if(cy[i] != -1 && dis == dy[i]) continue;
            if(cy[i] == -1 || dfs(cy[i]))
            {
                cy[i] = u;
                cx[u] = i;
                return 1;
            }
        }
    }
    return 0;
}

int hk()
{
    int res = 0;
    mem(cx, -1);
    mem(cy, -1);
    while(bfs())
    {
        mem(used, 0);
        for(int i=1; i<=nx; i++)
            if(cx[i] == -1 && dfs(i))
                res++;
    }
    return res;
}

int istwo(int u)
{
    queue<int> E;
    mem(vis, -1);
    E.push(u);
    vis[u] = 1;
    while(!E.empty())
    {
        u = E.front(); E.pop();
        for(int i=0; i<G[u].size(); i++)
        {
            int v = G[u][i];
            if(vis[v] == -1)
            {
                if(vis[u] == 1) vis[v] = 0;
                else vis[v] = 1;
                E.push(v);
            }
            else if(vis[v] == vis[u])
                return 0;
        }
    }
    return 1;
}



int main()
{
    while(cin>> n >> m && n+m)
    {
        for(int i=0; i<maxn; i++) G[i].clear();

        for(int i=0; i<m; i++)
        {
            int u, v;
            cin>> u >> v;
            G[u].push_back(v);
            G[v].push_back(u);
        }

        if(!istwo(1) || n == 1)
        {
            cout<< "No" <<endl;
            continue;
        }
        nx = n; ny = n;
        cout<< hk()/2 <<endl;

    }


    return 0;
}
View Code

 

The Accomodation of Students HDU - 2444(判断二分图 + 二分匹配)

标签:vector   style   using   main   output   problem   src   pre   close   

原文地址:https://www.cnblogs.com/WTSRUVF/p/9309629.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!