hdu 无题II（二分差值+最大匹配）

http://acm.hdu.edu.cn/showproblem.php?pid=2236

找n个数使得这n个数都在不同的行和列里显然是二分图模型。难点在于求最大值与最小值差值最小。这里二分差值(看的题解)，进行试探是否可以匹配成功。

#include <stdio.h>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#include <math.h>
#include <string.h>
#include <queue>
#define LL long long
#define _LL __int64
using namespace std;
const int INF = 0x3f3f3f3f;
const int mod = 1000000007;
const int maxn = 110;


int n;
int G[maxn][maxn];
int Max,Min;
int cur,mid;
int match[10010],chk[10010];

int dfs(int p)
{
	for(int i = 1; i <= n; i++)
	{
		if(!chk[i] && G[p][i] >= cur && G[p][i] <= cur+mid)
		{
			chk[i] = 1;
			if(match[i] == -1 || dfs(match[i]))
			{
				match[i] = p;
				return 1;
			}
		}
	}
	return 0;
}

int OK(int cur)
{
	memset(match,-1,sizeof(match));
	for(int i = 1; i <= n; i++)
	{
		memset(chk,0,sizeof(chk));
		if(!dfs(i))
			return 0;
	}
	return 1;
}

int main()
{
	int test;
	scanf("%d",&test);
	while(test--)
	{
		scanf("%d",&n);
		Max = -1;
		Min = 110;
		for(int i = 1; i <= n; i++)
		{
			for(int j = 1; j <= n; j++)
			{
				scanf("%d",&G[i][j]);
				if(G[i][j] > Max)
					Max = G[i][j];
				if(G[i][j] < Min)
					Min = G[i][j];
			}
		}

		int l,r;
		l = 0;
		r = Max - Min;
		int res;

		while(l <= r)
		{
			mid = (l+r) >> 1;
			int flag = 0;
			for(cur = Min; cur + mid <= Max; cur++) //枚举最小值，检查当前差值是否可以匹配成功
			{
				if( OK(cur) )
				{
					flag = 1;
					break;
				}
			}
			if(flag) //匹配成功,更新r
			{
				res = mid;
				r = mid-1;
			}
			else
				l = mid+1;//匹配不成功，更新l
		}
		printf("%d\n",res);
	}
	return 0;
}

hdu 无题II（二分差值+最大匹配）,布布扣,bubuko.com

hdu 无题II（二分差值+最大匹配）

原文：http://blog.csdn.net/u013081425/article/details/25309513