poj-1018-Communication System

Communication System

Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 28519 Accepted: 10150

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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

题意：有T组测试数据，每组1个n,表示n行，接下来n行，每行一个m，表示有m个管道，每个管道有流量和费用，最后求从n行中，每行选择1个管道，要求 B/P最大 ，B表示所选的那个方案中n个管道的最小的那个的流量，P表示n个管道费用和。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。

代码：

#include <stdio.h>#include <string.h>#define inf 9999999#define min(a,b) a<b?a:bint dp[120][1200];//前i个带宽容量为j的最小费用 int main()  {      int T;      scanf("%d",&T);      while(T--)      {          int n;          scanf("%d",&n);          for(int i=1; i<=n; i++)  //初始化          {              for(int j=0; j<1100; j++)                  dp[i][j]=inf;          }          for(int i=1; i<=n; i++)   //dp          {              int num;              scanf("%d",&num);              for(int j=1; j<=num; j++)              {                  int p,b;                  scanf("%d%d",&b,&p);                  if(i==1)                  {                      dp[1][b]=min(dp[1][b],p);                  }                  else                  {                      for(int k=0; k<1100; k++)                      {                          if(dp[i-1][k]!=inf)                          {                              if(k<=b)  //找最小                                dp[i][k]=min(dp[i][k],dp[i-1][k]+p);                              else                                  dp[i][b]=min(dp[i][b],dp[i-1][k]+p);                          }                      }                  }              }          }          double ans=0;          for(int i=0; i<1100; i++)          {              if(dp[n][i]!=inf)              {  //              printf("%d %d\n",i,dp[n][i]);                 double k=(double)i/dp[n][i];                  if(k>ans)                      ans=k;              }          }          printf("%.3lf\n",ans);      }      return 0;  }