题目：
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.
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.
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.
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。
思路：要使得B/P最大，那B应尽量大，而P要尽量小。由于B是n件物品中B值最小的一个，所以可以枚举n件配件的所有种类，每次枚举一种，把B值就定下来了使num=B，然后在剩余的n-1种配件中选取，而接下来选的配件的B值必须比num大，然后在这些符合条件的物品里算个P值最小的。当选好后，计算每次的B/P，找出最大的B/P。
代码：
#include<iostream>
#include<vector>
using namespace std;
int test,n,nums;
struct node
{
int b;
int p;
};
vector<node>q[100];
void Init()//清理q中的数据
{
int i;
for(i=0;i<100;i++)q[i].clear();
}
int main()
{
int i,j,k,m;
double max;
cin>>test;
node no1,no2;
while(test--)
{
Init();
cin>>n;
for(i=0;i<n;i++)
{
cin>>nums;
for(j=0;j<nums;j++)
{
cin>>no1.b>>no1.p;
q[i].push_back(no1);
}
}
max=0;
for(i=0;i<n;i++)
{
for(j=0;j<q[i].size();j++)
{
no1=q[i][j];
double sum=no1.p,num=no1.b;
int pp;
for(k=0;k<n;k++)
{
if(k==i)continue;
pp=999999;
for(m=0;m<q[k].size();m++)
{
no2=q[k][m];
if(no2.b<no1.b)continue;
if(pp>no2.p)pp=no2.p;
}
if(pp==999999)break;
sum+=pp;
}
if(pp==999999)break;
if(max<num/sum)max=num/sum;
}
}
printf("%.3f\n",max);
}
return 0;
}