POJ 1018

时间：2016-01-22 22:07:08 阅读：307 评论：0 收藏：0 [点我收藏+]

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Communication System

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 25744		Accepted: 9184

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

Sample Output

0.649

Source

Tehran 2002, First Iran Nationwide Internet Programming Contest

大致题意：

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。

同种设备都会存在两个方面的差别：bandwidths 和 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：

找出最大bandwidths和最小bandwidths，然后从小到大依次枚举，找出每种bandwidths的最大B/P（即最小prices），再比较出最大的即为answer；

 1 #include <iostream>
 2 #include <algorithm>
 3 using namespace std;
 4 #define N 105
 5 int main()
 6 {
 7     int a[N][N],b[N][N],tot[N];//bandwidths、prices 、输入计数 
 8     int t,n;
 9     scanf("%d",&t);
10     while(t--)
11     {
12         double ans=0;    
13         int MAX=0,MIN=0xffff;//最大最小 bandwidths
14         scanf("%d",&n);
15         for(int i=1;i<=n;i++)
16         {
17             scanf("%d",&tot[i]);
18             for(int j=1;j<=tot[i];j++)
19             {
20                 scanf("%d%d",&a[i][j],&b[i][j]);
21                 MAX=max(MAX,a[i][j]);
22                 MIN=min(MIN,a[i][j]);
23             }
24         }
25         for(int key=MIN;key<=MAX;key++)//枚举 
26         {
27             int sum=0;
28             for(int i=1;i<=n;i++)
29             {
30                 int M=0xffff;
31                 for(int j=1;j<=tot[i];j++)
32                     if(a[i][j]>=key&&M>b[i][j]) 
33                         M=b[i][j];
34                 sum+=M;
35             }
36         if(1.0*key/sum>ans) ans=1.0*key/sum;
37         }
38         printf("%.3lf\n",ans);
39     } return 0;
40 }

POJ 1018

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原文地址：http://www.cnblogs.com/nicetomeetu/p/5152291.html

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