HDU-2413 Against Mammoths

二分答案，对于当前答案Ans，求出某些人类可打败某些外星人的对应边，建图后求是否有完备匹配。

//#include <cmath>
#include <cstdlib>
#include <cstdio>
#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 259
#define MAX 1<<30
#define ll long long

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;
}

struct edge{int y, n;} e[N*N]; int fir[N], en;
int na, nb, ma[N], mb[N], ra[N], rb[N], now, k[N], b[N], d[N][N], l, r;
ll t;

void Add(int x, int y) { en++, e[en].y=y, e[en].n=fir[x], fir[x]=en; }

bool Find(int x)
{
	int o=fir[x], y=e[o].y;
	while (o)
	{
		if (b[y]!=now)
		{
			b[y]=now;
			if (!k[y] || Find(k[y])) { k[y]=x; return true; }
		}
		o=e[o].n, y=e[o].y;
	}
	return false;
}

int main()
{
	while(scanf("%d %d", &na, &nb) && (na || nb))
	{
		rep(i, 1, na) ma[i]=read(), ra[i]=read();
		rep(i, 1, nb) mb[i]=read(), rb[i]=read();
		rep(i, 1, na) rep(j, 1, nb) d[i][j]=read();

		l=0, r=MAX;
		while (l!=r)
		{
			t=(l+r)/2; bool can=true;
			rep(i, 1, nb) fir[i]=0; en=0;
			rep(i, 1, na)
				rep(j, 1, nb)
				{
					if(d[i][j] > t) continue;
					ll s1 = ma[i]+ra[i]*(t-d[i][j]);
					ll s2 = mb[j]+rb[j]*t;
					if(s1 >= s2) Add(j, i);
				}
			rep(i, 1, na) k[i]=b[i]=0;
			rep(i, 1, nb) if (!Find(now=i)) { can=false; break; }
			if (can) r=t; else l=t+1;
		}
		if (l==MAX) printf("IMPOSSIBLE\n"); else printf("%d\n", l);
	}
	return 0;
}