poj1018 枚举

http://poj.org/problem?id=1018

Description

We have received an order from Pizoor Communications Inc. for a special communication system. The system consists of several devices. For each device, we are free to choose from several manufacturers. Same devices from two manufacturers differ in their maximum bandwidths and prices. 
By overall bandwidth (B) we mean the minimum of the bandwidths of the chosen devices in the communication system and the total price (P) is the sum of the prices of all chosen devices. Our goal is to choose a manufacturer for each device to maximize B/P. 

Input

The first line of the input file contains a single integer t (1 ≤ t ≤ 10), the number of test cases, followed by the input data for each test case. Each test case starts with a line containing a single integer n (1 ≤ n ≤ 100), the number of devices in the communication system, followed by n lines in the following format: the i-th line (1 ≤ i ≤ n) starts with mi (1 ≤ mi ≤ 100), the number of manufacturers for the i-th device, followed by mi pairs of positive integers in the same line, each indicating the bandwidth and the price of the device respectively, corresponding to a manufacturer.

Output

Your program should produce a single line for each test case containing a single number which is the maximum possible B/P for the test case. Round the numbers in the output to 3 digits after decimal point. 

Sample Input

1 33 100 25 150 35 80 252 120 80 155 402 100 100 120 110

Sample Output

0.649

/**poj 1018  枚举题目大意：某公司要建立一套通信系统，该通信系统需要n种设备，而每种设备分别可以有m1、m2、m3、...、mn个厂家提供生产，而每个厂家生产的同种设备都会存在两个方面的差别：带宽bandwidths 和 价格prices。          现在每种设备都各需要1个，考虑到性价比问题，要求所挑选出来的n件设备，要使得B/P最大。其中B为这n件设备的带宽的最小值，P为这n件设备的总价。解题思路：这道题的a[i][j]较小，可以用枚举直接过，枚举所有可能的B值。（所有组别中最小的最小值和最小的最大值之间，至于flag判断没有也可以，只是这样可以优化一下时间复杂度）                    */#include <stdio.h>#include <string.h>#include <iostream>using namespace std;struct note{    int x,y;} a[105][105];int n,num[105],maxx[105],minn[105],flag[100005];int find(int b){    int ret=0;    for(int i=0; i<n; i++)    {        int minp=0x3f3f3f3f;        for(int j=0; j<num[i]; j++)        {            if(a[i][j].x>=b)                minp=min(minp,a[i][j].y);        }        ret+=minp;    }    return ret;}int main(){    int T;    scanf("%d",&T);    while(T--)    {        scanf("%d",&n);        memset(flag,0,sizeof(flag));        for(int i=0; i<n; i++)        {            scanf("%d",&num[i]);            for(int j=0; j<num[i]; j++)            {                scanf("%d%d",&a[i][j].x,&a[i][j].y);                flag[a[i][j].x]=1;            }        }        for(int i=0; i<n; i++)        {            minn[i]=maxx[i]=a[i][0].x;            for(int j=1; j<num[i]; j++)            {                maxx[i]=max(maxx[i],a[i][j].x);                minn[i]=min(minn[i],a[i][j].x);            }        }        int s=minn[0];        int e=maxx[0];        for(int i=1; i<n; i++)        {            s=min(s,minn[i]);            e=min(e,maxx[i]);        }        double ans=0;        for(int i=s; i<=e; i++)        {            if(flag[i])            {                int d=find(i);                ans=max(ans,i*1.0/d);            }        }        printf("%.3lf\n",ans);    }    return 0;}