zoukankan      html  css  js  c++  java
  • HDU2639——背包DP(K最优解)——Bone Collector II

    Description

    The title of this problem is familiar,isn't it?yeah,if you had took part in the "Rookie Cup" competition,you must have seem this title.If you haven't seen it before,it doesn't matter,I will give you a link: 

    Here is the link: http://acm.hdu.edu.cn/showproblem.php?pid=2602

    Today we are not desiring the maximum value of bones,but the K-th maximum value of the bones.NOTICE that,we considerate two ways that get the same value of bones are the same.That means,it will be a strictly decreasing sequence from the 1st maximum , 2nd maximum .. to the K-th maximum. 

    If the total number of different values is less than K,just ouput 0.
     

    Input

    The first line contain a integer T , the number of cases. 
    Followed by T cases , each case three lines , the first line contain two integer N , V, K(N <= 100 , V <= 1000 , K <= 30)representing the number of bones and the volume of his bag and the K we need. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone. 
     

    Output

    One integer per line representing the K-th maximum of the total value (this number will be less than 2 31). 
     

    Sample Input

    3 5 10 2 1 2 3 4 5 5 4 3 2 1 5 10 12 1 2 3 4 5 5 4 3 2 1 5 10 16 1 2 3 4 5 5 4 3 2 1
     

    Sample Output

    12 2 0
     大意:背包问题的最优解问题
    用暴力
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    int main()
    {
        int w[1005],v[1005];
        int t1[1005],t2[1005];
        int dp[1005][105];
        int n,m,k,T;
        scanf("%d",&T);
        while(T--){
            scanf("%d%d%d",&n,&m,&k);
            for(int i = 1; i <= n ; i ++)
                scanf("%d",&v[i]);
            for(int i = 1; i <= n; i ++)
                scanf("%d",&w[i]);
           memset(dp,0,sizeof(dp));
           if(k == 0){
               printf("0
    ");
           continue;
           }
        for(int i = 1;i <= n ; i++){
            for(int j = m; j >= w[i];j--){
                for(int l = 1; l <= k ; l++){
                    t1[l] = dp[j][l];
                    t2[l] = dp[j-w[i]][l]+v[i];
                }
                int x = 1,y = 1, t = 1;
                t1[k+1] = t2[k+1] = -1;
                while(t <= k && ( x <= k || y <= k)){
                    if(t1[x] > t2[y])
                    dp[j][t] = t1[x++];
                    else 
                    dp[j][t] = t2[y++];
                    if(t == 1 || dp[j][t]!=dp[j][t-1])
                        t++;
                }
            }
        }
        printf("%d
    ",dp[m][k]);
      }
    return 0;
    }
    View Code
  • 相关阅读:
    linux正则表达式
    linux监控值free命令详解
    loadrunner 中数组的定义
    管线工具sed和正则表达式
    web_link()参数化
    mysql执行插入时性能优化
    linux关闭防火墙
    linux系统中变量和环境变量
    mysql源码安装与基本优化配置
    Memory Consistency Erros
  • 原文地址:https://www.cnblogs.com/zero-begin/p/4448515.html
Copyright © 2011-2022 走看看