【HDU2444】【匈牙利算法】【二分匹配】【求最大匹配】【染色】

The Accomodation of Students

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3690    Accepted Submission(s): 1730

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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#include <iostream>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <string>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
using namespace std;

#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define mp push_back

int n,m;
const int N = 220;
const int M = 220 * 220;
struct Edge
{
	int to,next;
}edge[M];
int len;
int H[N];
int col[N];
int match[N*2];
bool used[N];

void addedge(int u,int v)
{
	edge[len].to = v;
	edge[len].next = H[u];
	H[u] = len ++;
}
void init()
{
	len = 0;
	memset(H,-1,sizeof(H));
	memset(col,0,sizeof(col));
	memset(match,-1,sizeof(match));
	memset(used,0,sizeof(used));
}

bool colorit(int i,int c)
{
	col[i] = c;
	for(int ei=H[i];~ei;ei=edge[ei].next)
	{
		int v = edge[ei].to;
		if(col[v] == 0)
		{
			if(!colorit(v,-c))
				return false;
		}
		else if(col[v] == c)
		{
			return false;
		}
	}
	return true;
}
bool ok(int n)
{
	for(int i=1;i<=n;i++)
	{
		if(col[i] == 0)
		{
			if(!colorit(i,1))
				return false;
		}
	}
	return true;
}

bool dfs(int u)
{
	used[u] = true;
	for(int ei=H[u];ei!=-1;ei=edge[ei].next)
	{
		int v = edge[ei].to;
		
		if(match[v] == -1 || !used[match[v]] && dfs(match[v]))
		{
			//match[v] = u;
			match[u] = v;
			return true;
		}
	}
	return false;

}
void solve(int n)
{
	if(!ok(n))
	{
		puts("No");
		return;
	}
	int ret = 0;
	for(int i=1;i<=n;i++){
		if(match[i] == -1)
		{
			memset(used,0,sizeof(used));
			if(dfs(i)) ret ++;
		}
	}
	printf("%d\n",ret);

}
int main()
{
	while(scanf("%d%d",&n,&m) != EOF)
	{
		int u,v;
		//cout << n << m << endl;
		init();
		for(int i=0;i<m;i++) {
			scanf("%d%d",&u,&v);
			addedge(u,v);
			addedge(v,u);
		}

		solve(n);
	}
	return 0;
}