poj 1018 Communication System_贪心

题意：给你n个厂，每个厂有m个产品，产品有B（带宽），P（价格），现在要你求最大的 B/P

明显是枚举，当P大于一定值，B/P为零，可以用这个剪枝

#include <iostream>
#include<cstring>
#include<cstdio>
using namespace std;
#define N 110
#define INF 0xffffff
int devb[N][N],devp[N][N];
int b[N*100],tb;

int main(int argc, char** argv) {
	int n,mi[N],i,j,t;
	int blen,minprice,sum,minb,maxb,curb;
	double dmax,tvalue;
	scanf("%d",&t);
	while(t--){
		scanf("%d",&n);
		blen=0;
		for(i=0;i<n;i++){
			scanf("%d",&mi[i]);
			for(j=0;j<mi[i];j++){
				scanf("%d%d",&devb[i][j],&devp[i][j]);
				b[blen++]=devb[i][j];
				if(minb>=devb[i][j])
					minb=devb[i][j];
				if(maxb<=devb[i][j])
					maxb=devb[i][j];
			}
		}
		dmax=0;
		for(tb=minb;tb<=maxb;tb++){
			sum=0;
			curb=INF;
			for(i=0;i<n;i++){
				minprice=INF;
				for(j=0;j<mi[i];j++){
					if(devb[i][j]>=tb&&devp[i][j]<minprice){
						minprice=devp[i][j];
						if(curb>devb[i][j])
							curb=devb[i][j];
					}
				}
				if(minprice==INF)
					break;
				sum+=minprice;
			}
			if(i==n){
				tvalue=(double)curb/sum;
				if(tvalue>dmax)
					dmax=tvalue;
				tb=curb+1;
			}else
				break;
		}
		printf("%.3f\n",dmax);
	}
	return 0;
}