ZOJ 1409 Communication System（枚舉 + 貪心）

Communication System

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Time Limit: 10 Seconds Memory Limit: 32768 KB

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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
3
3 100 25 150 35 80 25
2 120 80 155 40
2 100 100 120 110

Sample Output

0.649

題意：有n種設備，每種有mi件，每件設備有一個帶寬和價格，從每種設備中選一件，使得帶寬B/總價格P的值最大。B為挑選的n件設備中帶寬最小的設備的帶寬，P為n件設備的總價格。 分析：要求最大的帶寬價值比，只需讓帶寬盡可能的大，總價格盡可能的小就行。所以可以枚舉所有可能的帶寬，然後從選出n件設備，每種設備貪心選擇滿足帶寬限制的價格最小的，這樣能保證總價格最小。

#include
#include
#include
using namespace std;
set  s; //保存所有可能的帶寬
set  ::iterator it;
int band[105][105], price[105][105], m[105];
int main() {
    int T, n, i, j;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        s.clear();
        for(i = 0; i < n; i++)
        {
            scanf("%d",&m[i]);
            for(j = 0; j < m[i]; j++)
            {
                scanf("%d%d",&band[i][j], &price[i][j]);
                s.insert(band[i][j]);
            }
        }
        double rate = 0;
        for(it = s.begin(); it != s.end(); it++)  //枚舉帶寬
        {
            int k = *it, sum_price = 0;
            for(i = 0; i < n; i++) //枚舉第i種設備
            {
                int min_price = 0x3fffff;
                for(j = 0; j < m[i]; j++) //枚舉第i種設備的第j件
                {
                    if(band[i][j] >= k && price[i][j] < min_price) //挑選滿足條件的價格最小的
                    {
                        min_price = price[i][j];
                    }
                }
                sum_price += min_price;
            }
            if(k * 1.0 / sum_price > rate)
                rate = k * 1.0 / sum_price;
        }
        printf("%.3lf\n",rate);
    }
    return 0;
}