zoukankan      html  css  js  c++  java
  • UVA624——背包DP(回溯)——CD

    Description

     

    You have a long drive by car ahead. You have a tape recorder, but unfortunately your best music is on CDs. You need to have it on tapes so the problem to solve is: you have a tape N minutes long. How to choose tracks from CD to get most out of tape space and have as short unused space as possible.


    Assumptions:

    • number of tracks on the CD. does not exceed 20
    • no track is longer than N minutes
    • tracks do not repeat
    • length of each track is expressed as an integer number
    • N is also integer

    Program should find the set of tracks which fills the tape best and print it in the same sequence as the tracks are stored on the CD

    Input 

    Any number of lines. Each one contains value N, (after space) number of tracks and durations of the tracks. For example from first line in sample data: N=5, number of tracks=3, first track lasts for 1 minute, second one 3 minutes, next one 4 minutes

    Output 

    Set of tracks (and durations) which are the correct solutions and string `` sum:" and sum of duration times.

    Sample Input 

    5 3 1 3 4
    10 4 9 8 4 2
    20 4 10 5 7 4
    90 8 10 23 1 2 3 4 5 7
    45 8 4 10 44 43 12 9 8 2
    

    Sample Output 

    1 4 sum:5
    8 2 sum:10
    10 5 4 sum:19
    10 23 1 2 3 4 5 7 sum:55
    4 10 12 9 8 2 sum:45

     大意:最优解的回溯.要求输出最优解的各个阶段的值,用一个path[i][j]记录(PS:偶尔用个index结果是库定义~~~~编译错了四次....),因为输出是从前到后,一开始就要最优解,所以前面01背包的处理要反一下,每一次的得到最优解之后,j-=w[i]

    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    int main()
    {
        int dp[10005],w[25];
        int path[25][10005];
        int n,m;
        while(~scanf("%d%d",&n,&m)){
            memset(dp,0,sizeof(dp));
            memset(w,0,sizeof(w));
            memset(path,0,sizeof(path));
            for(int i =1; i <= m ; i++)
                scanf("%d",&w[i]);
            for(int i = m; i >= 1 ;i--){
                for(int j = n; j >= w[i]; j--){
                    if(dp[j] < dp[j-w[i]]+w[i]){
                        dp[j] = dp[j-w[i]]+w[i];
                        path[i][j] = 1;
                    }
                }
             }
            for(int i = 1,j = n; i <= m ; i++){
                if(path[i][j]){
               printf("%d ",w[i]);
                j -= w[i];
                }
            }
           printf("sum:%d
    ",dp[n]);
        }
       return 0;
    } 
    View Code
     
     
  • 相关阅读:
    window 7 安装Jmeter并配置https录制脚本
    windows 7安装Fiddler抓HTTPS请求的解决办法
    APP界面设计与页面布局的23条基本原则
    Charles 连接手机抓包出现Unknown,一直无法抓包的问题解决
    Charles 如何破解与连接手机进行抓包
    WeTest----如何查看Wetest生成测试报告?
    WeTest----如何使用WeTest进行App性能测试?
    兼容位图
    DeleteDC ReleaseDC DeleteObject之间的区别
    屏幕保存为位图
  • 原文地址:https://www.cnblogs.com/zero-begin/p/4445565.html
Copyright © 2011-2022 走看看