zoukankan      html  css  js  c++  java
  • Accepted Necklace

    Accepted Necklace
    Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 3136 Accepted Submission(s): 1213

    Problem Description
    I have N precious stones, and plan to use K of them to make a necklace for my mother, but she won’t accept a necklace which is too heavy. Given the value and the weight of each precious stone, please help me find out the most valuable necklace my mother will accept.

    Input
    The first line of input is the number of cases.
    For each case, the first line contains two integers N (N <= 20), the total number of stones, and K (K <= N), the exact number of stones to make a necklace.
    Then N lines follow, each containing two integers: a (a<=1000), representing the value of each precious stone, and b (b<=1000), its weight.
    The last line of each case contains an integer W, the maximum weight my mother will accept, W <= 1000.

    Output
    For each case, output the highest possible value of the necklace.

    Sample Input

    1
    2 1
    1 1
    1 1
    3

    Sample Output

    1

    背包问题,DP[i][m][k]表示选第i件物品时的体积为m,以选定的物品为k件,所以Dp[i][m][k]=max(Dp[i-1][m][k],Dp[i-1][m-v[i]][k-1]+w[i]);再将其转化为01背包

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <cmath>
    #include <queue>
    #include <algorithm>
    using namespace std;
    typedef long long LL;
    typedef pair<int,int>p;
    const int MAX =  1100;
    int Dp[MAX][35];
    p Th[35];
    int main()
    {
        int n,k,m;
        int T;
        scanf("%d",&T);
        while(T--)
        {
            scanf("%d %d",&n,&k);
            for(int i=1;i<=n;i++)
            {
                scanf("%d %d",&Th[i].first,&Th[i].second);
            }
            scanf("%d",&m);
            memset(Dp,0,sizeof(Dp));
            for(int i=1;i<=n;i++)
            {
                for(int j=m;j>=Th[i].second;j--)
                {
                    for(int s=1;s<=k;s++)
                    {
                        Dp[j][s]=max(Dp[j-Th[i].second][s-1]+Th[i].first,Dp[j][s]);
                    }
                }
            }
            printf("%d
    ",Dp[m][k]);
        }
        return 0;
    }
    
  • 相关阅读:
    Linux设备模型 学习总结
    平衡二叉树
    数字在排序数组中出现的次数
    两个链表的第一个公共节点
    第一个只出现一次的字符
    丑数
    把数组排成最小的数
    剑指offer 连续子数组的最大和
    查找描述信息中包括robot的电影对应的分类名称以及电影数目,而且还需要该分类对应电影数量>=5部
    for each
  • 原文地址:https://www.cnblogs.com/juechen/p/5255976.html
Copyright © 2011-2022 走看看