HDU-2236 无题II

选取不同列不同行的N个数。。。明摆着叫你二分匹配

二分答案,然后枚举边的范围并跑匈牙利,以此判断答案范围。

#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <fstream>
#include <iostream>

#define rep(i, l, r) for(int i=l; i<=r; i++)
#define down(i, l, r) for(int i=l; i>=r; i--)
#define N 123

using namespace std;
int read()
{
	int x=0, f=1; char ch=getchar();
	while (ch<'0' || ch>'9') { if (ch=='-') f=-1; ch=getchar(); }
	while (ch>='0' && ch<='9') { x=x*10+ch-'0'; ch=getchar(); }
	return x*f;
}

int n, l, r, m, low, d[N][N], k[N];
bool b[N];

bool Find(int x)
{
	rep(i, 1, n) if (d[x][i]>=low && d[x][i]<=low+m && !b[i])
	{
		b[i]=1; if (!k[i] || Find(k[i])) { k[i]=x; return true; }
	}
	return false;
}

int main()
{
	int t=read(); while (t--)
	{
		
		n=read();
		rep(i, 1, n) rep(j, 1, n) d[i][j]=read();
		int l=0, r=100;
		while (l!=r)
		{
			m=(l+r)/2; low=0; bool can=false;
			while (low+m<=100)
			{
				bool can2=true;
				rep(i, 1, n) k[i]=0;
				rep(i, 1, n)
				{
					rep(j, 1, n) b[j]=0;
					if (!Find(i)) { can2=false; break; }
				}
				if (can2) { can=true; break; }
				low++;
			}
			if (can) r=m; else l=m+1;
		}
		printf("%d
", l);
	}
	return 0;
}
原文地址:https://www.cnblogs.com/NanoApe/p/4345563.html