POJ 1018 - Communication System

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

大致题意：要建立一个系统需要n种设备，每种设备由m个公司提供，再分别给出每个m的带宽和价格。

简单来说就是一个01分组背包问题，但是不简单，需要先找出满足要求带宽中对应的最小价格，再从 (带宽)/(价格) 中找到最大值即可。

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int ans[120][1200];///ans[i][j]选择了前 i 个带宽是 j 的最小费用const int INF = 0x3f3f3f3f;int main(){    int T;    int n;    int num;    scanf("%d", &T);    while (T--)    {        scanf("%d", &n);        for (int i = 1; i <= n; ++i)///初始化        {            for (int j = 0; j < 1100; ++j)                ans[i][j] = INF;        }        int brand, price;        for (int i = 1; i <= n; ++i)        {            scanf("%d", &num);            for (int j = 1; j <= num; ++j)            {                scanf("%d%d", &brand, &price);                if (i == 1)                    ans[1][brand] = min(ans[1][brand], price);                else                {                    for (int k = 0; k < 1100; ++k)                    {                        if (ans[i-1][k] != INF)                        {                            if (k <= brand)                                ans[i][k] = min(ans[i][k], ans[i-1][k]+price);                            else                                ans[i][brand] = min(ans[i][brand], ans[i-1][k]+price);                        }                    }                }            }        }        double ret = 0.0;        for (int i = 0; i < 1100; ++i)        {            if (ans[n][i] != INF)            {                double tmp = (double)i/ans[n][i]*1.0;                //printf("%lf\n", tmp);                if (tmp > ret)                    ret = tmp;            }        }        printf("%.3lf\n", ret);    }    return 0;}