BOJ 6399 奔小康赚大钱 //二分图KM模板题

约定不等于承诺〃 2022-08-10 14:45 232阅读 0赞

# 奔小康赚大钱 #

Time Limit: 1000ms

Memory Limit: 32768KB

This problem will be judged on HDU. Original ID: [2255][]  
64-bit integer IO format: %I64d Java class name: Main

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传说在遥远的地方有一个非常富裕的村落,有一天,村长决定进行制度改革：重新分配房子。  
这可是一件大事,关系到人民的住房问题啊。村里共有n间房间,刚好有n家老百姓,考虑到每家都要有房住（如果有老百姓没房子住的话，容易引起不安定因素），每家必须分配到一间房子且只能得到一间房子。  
另一方面,村长和另外的村领导希望得到最大的效益,这样村里的机构才会有钱.由于老百姓都比较富裕,他们都能对每一间房子在他们的经济范围内出一定的价格,比如有3间房子,一家老百姓可以对第一间出10万,对第2间出2万,对第3间出20万.(当然是在他们的经济范围内).现在这个问题就是村领导怎样分配房子才能使收入最大.(村民即使有钱购买一间房子但不一定能买到,要看村领导分配的).

## Input ##

输入数据包含多组测试用例，每组数据的第一行输入n,表示房子的数量(也是老百姓家的数量)，接下来有n行,每行n个数表示第i个村名对第j间房出的价格(n<=300)。

## Output ##

请对每组数据输出最大的收入值，每组的输出占一行。

## Sample Input ##

2
    100 10
    15 23

## Sample Output ##

123

## Source ##

[HDOJ 2008 Summer Exercise（4）- Buffet Dinner][HDOJ 2008 Summer Exercise_4_- Buffet Dinner]

#include <stdio.h>
    #include <string.h>
    
    #define p 305
    #define m 999999999
    
    int n, tu[p][p];
    int link[p], slack[p];
    int lx[p], ly[p];
    int visitx[p], visity[p];
    
    int MAX(int a, int b)
    {
    	if(a>b)
    		return a;
    	return b;
    }
    int MIN(int a, int b)
    {
    	if(a<b)
    		return a;
    	return b;
    }
    
    int dfs(int k)
    {
    	int i, d;
    
    	visitx[k] = 1;
    	for(i=0; i<n; i++)
    	{
    		if(visity[i])
    			continue;
    		d = lx[k]+ly[i]-tu[k][i];
    
    		if(d==0)
    		{
    			visity[i] = 1;
    			if(link[i]==-1 || dfs(link[i]))
    			{
    				link[i] = k;
    				return 1;
    			}
    		}
    		else
    			slack[i] = MIN(slack[i], d);
    	}
    	return 0;
    }
    int KM()
    {
    	int i, j, ans;
    
    	for(i=0; i<n; i++)
    		link[i] = -1;
    	memset(lx, 0, sizeof(lx));
    	memset(ly, 0, sizeof(ly));
    	for(i=0; i<n; i++)
    	{
    		for(j=0; j<n; j++)
    			lx[i] = MAX(lx[i], tu[i][j]);
    	}
    	for(i=0; i<n; i++)
    	{
    		for(j=0; j<n; j++)
    		{
    			slack[j] = m;
    		}
    		while(1)
    		{
    			memset(visitx, 0, sizeof(visitx));
    			memset(visity, 0, sizeof(visity));
    			if(dfs(i))
    				break;
    			else
    			{
    				int d;
    				d = m;
    				for(j=0; j<n; j++)
    				{
    					if(!visity[j])
    						d = MIN(d, slack[j]);
    				}
    				for(j=0; j<n; j++)
    				{
    					if(visitx[j])
    						lx[j] -= d;
    					if(visity[j])
    						ly[j] += d;
    				}
    			}
    		}
    	}
    	ans = 0;
    	for(i=0; i<n; i++)
    		ans += tu[link[i]][i];
    	return ans;
    }
    int main()
    {
    	int i, j;
    	while(scanf("%d", &n)!=-1)
    	{
    		for(i=0; i<n; i++)
    		{
    			for(j=0; j<n; j++)
    			{
    				scanf("%d", &tu[i][j]);
    			}
    		}
    		printf("%d\n", KM());
    	}
    	return 0;
    }

[2255]: http://acm.hdu.edu.cn/showproblem.php?pid=2255
[Link 1]: http://www.bnuoj.com/bnuoj/problem_show.php?pid=6399#
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[HDOJ 2008 Summer Exercise_4_- Buffet Dinner]: http://www.bnuoj.com/bnuoj/problem.php?search=HDOJ+2008+Summer+Exercise%EF%BC%884%EF%BC%89-+Buffet+Dinner