HDOJ 2444 The Accomodation of Students

最新推荐文章于 2021-09-14 21:40:10 发布

转载最新推荐文章于 2021-09-14 21:40:10 发布 · 48 阅读

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原文链接：http://www.cnblogs.com/CKboss/p/3350832.html

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#java

本博客探讨了如何将学生群体分为两组，使得同一组内的学生互不认识，并利用匈牙利算法解决住宿安排问题，以实现最大数量的学生入住双人间。

染色判读二分图+Hungary匹配

The Accomodation of Students

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1705 Accepted Submission(s): 821

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

2008 Asia Harbin Regional Contest Online

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gaojie

#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>

using namespace std;

typedef struct
{
    int to,next;
}Edge;

Edge E[50000];
int Adj[50000],Size=0,color[500],n,m,from[500];
bool use[500];

void Init()
{
    Size=0;
    memset(Adj,-1,sizeof(Adj));
}

void Add_Edge(int u,int v)
{
    E[Size].to=v;
    E[Size].next=Adj;
    Adj=Size++;
}

bool match(int x)
{
    for(int i=Adj[x];~i;i=E.next)
    {
        int v=E.to;
        if(!use[v])
        {
            use[v]=true;
            if(from[v]==-1||match(from[v]))
            {
                from[v]=x;
                return true;
            }
        }
    }
    return false;
}

int hungary()
{
    int sum=0;
    memset(from,-1,sizeof(from));
    for(int i=1;i<=n;i++)
    {
        if(color==1)
        {
            memset(use,false,sizeof(use));
            sum+=match(i);
        }
    }
    return sum;
}

int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        int u,v;
        Init();
        while(m--)
        {
            scanf("%d%d",&u,&v);
            Add_Edge(u,v);
            Add_Edge(v,u);
        }
        memset(color,0,sizeof(color));
        bool flag=true;
while(true)
{
        int pos=-1;
        if(flag==false) break;
        for(int i=1;i<=n;i++)
        {
            if(Adj==-1) continue;
            else if(color!=0) continue;
            else
            {
                pos=i; break;
            }
        }
        if(pos==-1) break;
        queue<int> q;
        q.push(pos); color[pos]=1;
        //printf("%d: \n",pos);
        while(!q.empty())
        {
            if(flag==false) break;
            int u=q.front(); q.pop();
            for(int i=Adj;~i;i=E.next)
            {
                int v=E.to;
                //printf("%d--->%d\n",u,v);
                if(color[v]==0)
                {
                    color[v]=-color;
                    q.push(v);
                }
                else if(color[v]==color)
                {
                    flag=false; break;
                }
            }
        }
}
        if(flag==false)
        {
            puts("No");
        }
        else
        {
            printf("%d\n",hungary());
        }
    }
    return 0;
}