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.
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 3 3 100 25 150 35 80 25 2 120 80 155 40 2 100 100 120 110
Sample Output
0.649
/*********************************************************************************************************************** 题意：某公司要买n个设备，每个设备都有不同的生产商，每个生产商的带宽b和价格p不同，要求选取的设备中的带宽最小的设备 b/买的所有设备的价格的总和p最大 思路：要使b/p最大，则b尽可能的大,p尽可能的小 首先我们选一个设备b出来，认为它的b最小 则其他设备的b小于它的全部跳过，再在这些设备中选取价格最小的即可 ***********************************************************************************************************************/ #include <cstdio> #include <cstring> #include <cmath> #include <queue> #include <stack> #include <vector> using namespace std; int min(int a ,int b) { return a > b ? b : a; } double max(double a, double b) { return a > b ? a : b; } typedef struct { int b, p; }node; vector <node> a[100+10]; int n , m , mina , sum; double ans = 0; node temp; void work() { for(int i = 0 ; i < n ; i ++) { for(int j = 0 ; j < a[i].size() ; j ++) { temp = a[i][j]; sum = temp.p; for(int k = 0 ; k < n ; k ++) { if(k == i) continue; mina = 0x7fffffff; for(int q = 0 ; q < a[k].size(); q++) { if(temp.b > a[k][q].b) continue; mina = min(mina , a[k][q].p); } if(mina == 0x7fffffff) break; sum += mina; } if(mina == 0x7fffffff) break; ans = max(ans , (double)temp.b / (double)sum); } } } void clr() { for(int i = 0 ; i < 100 + 5 ; i ++) a[i].clear(); } int main() { //freopen("data.in" , "r" , stdin); int t; scanf("%d", &t); while(t--) { ans = 0; clr(); scanf("%d", &n); for(int i = 0 ; i < n ; i ++) { scanf("%d", &m); for(int j = 0 ; j < m ; j ++) { node tm; scanf("%d %d", &tm.b, &tm.p); a[i].push_back(tm); } } work(); printf("%.3f " , ans); } return 0; }