zoukankan      html  css  js  c++  java
  • poj 1018(dp)

    Communication System
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 25653   Accepted: 9147

    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

    思路:定义dp[i][j]为前i个设备的容量为j的最小费用;
        状态转移方程为:dp[i][j]=min(dp[i][j],dp[i-1][j]+p);
        边界:dp[1][j]=p;  
        

      
     1 #include<iostream>
     2 #include<cstdio>
     3 #include<algorithm>
     4 #include<cstring>
     5 #include<cstdlib>
     6 #include<iomanip>
     7 #include<cmath>
     8 #include<vector>
     9 #include<queue>
    10 #include<stack>
    11 using namespace std;
    12 #define PI 3.141592653589792128462643383279502
    13 const int inf=0x3f3f3f3f;
    14 int t,n,j,ss,m[120][1200];
    15 int main(){
    16     //#ifdef CDZSC_June
    17     //freopen("in.txt","r",stdin);
    18     //#endif
    19     //std::ios::sync_with_stdio(false);
    20     scanf("%d",&t);
    21     while(t--){
    22         scanf("%d",&n);
    23         memset(m,0x3f,sizeof(m));
    24         for(int i=1;i<=n;i++){
    25             int num;
    26             scanf("%d",&num);
    27             for(j=1;j<=num;j++){
    28                 int b,p;
    29                 scanf("%d%d",&b,&p);
    30                 if(i==1){m[1][b]=min(m[1][b],p);}
    31                 else {
    32                     for(int k=0;k<1200;k++){
    33                         if(m[i-1][k]!=inf){
    34                             if(k<=b)
    35                                 m[i][k]=min(m[i][k],m[i-1][k]+p);
    36                             else
    37                                 m[i][b]=min(m[i][b],m[i-1][k]+p);
    38                         }
    39                     }
    40                 }
    41             }
    42         }
    43     double ans=0;
    44         for(int i=0;i<1200;i++){
    45             if(m[n][i]!=inf){
    46                 double k=(double)i/m[n][i];
    47                 if(k>ans) ans=k;
    48             }
    49         }
    50         printf("%.3lf
    ",ans);
    51     }
    52     return 0;
    53 }
  • 相关阅读:
    python爬取代理IP地址
    神经网络训练的过程
    机器学习中用到的数学概念
    Navicat连接Mysql错误代码1251
    mysql安装
    mysql运行找不到MSVCP140.dll
    tomcat 日志乱码
    扁平化 Flat
    常见的WEB安全及防护
    CentOS ceph 集群搭建(单节点)
  • 原文地址:https://www.cnblogs.com/yoyo-sincerely/p/5092736.html
Copyright © 2011-2022 走看看