zoukankan      html  css  js  c++  java
  • HDU3466背包01

    Proud Merchants

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
    Total Submission(s): 3126    Accepted Submission(s): 1288


    Problem Description
    Recently, iSea went to an ancient country. For such a long time, it was the most wealthy and powerful kingdom in the world. As a result, the people in this country are still very proud even if their nation hasn’t been so wealthy any more.
    The merchants were the most typical, each of them only sold exactly one item, the price was Pi, but they would refuse to make a trade with you if your money were less than Qi, and iSea evaluated every item a value Vi.
    If he had M units of money, what’s the maximum value iSea could get?

     

     

    Input
    There are several test cases in the input.

    Each test case begin with two integers N, M (1 ≤ N ≤ 500, 1 ≤ M ≤ 5000), indicating the items’ number and the initial money.
    Then N lines follow, each line contains three numbers Pi, Qi and Vi (1 ≤ Pi ≤ Qi ≤ 100, 1 ≤ Vi ≤ 1000), their meaning is in the description.

    The input terminates by end of file marker.

     

     

    Output
    For each test case, output one integer, indicating maximum value iSea could get.

     

     

    Sample Input
    2 10 10 15 10 5 10 5 3 10 5 10 5 3 5 6 2 7 3
     

     

    Sample Output
    5 11
     
     
     
     
     
     
     
     
     
    #include<stdio.h>
    #include<string.h>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    int dp[5005];
    int p[505],q[505],c[505];
    int n,m;
    int pd1(int p[],int q[],int c[])
    {
        int temp;
        for(int i=1;i<=n-1;i++)
            for(int j=i+1;j<=n;j++)
        {
            if(q[i]-p[i]>q[j]-p[j])
            {
                temp=p[i];
                p[i]=p[j];
                p[j]=temp;
                temp=q[i];
                q[i]=q[j];
                q[j]=temp;
                temp=c[i];
                c[i]=c[j];
                c[j]=temp;
            }
        }
    }
    int main()
    {
        while(scanf("%d%d",&n,&m)!=EOF)
        {
            memset(p,0,sizeof(p));
            memset(q,0,sizeof(q));
            memset(c,0,sizeof(c));
            memset(dp,0,sizeof(dp));
            for(int k=1;k<=n;k++)
                scanf("%d%d%d",&p[k],&q[k],&c[k]);
                pd1(p,q,c);
                for(int i=1;i<=n;i++)
                  for(int j=m;j>=q[i];j--)
                    dp[j]=max(dp[j],dp[j-p[i]]+c[i]);
                  printf("%d ",dp[m]);
        }
        return 0;
    }
  • 相关阅读:
    动词 + to do、动词 + doing
    图像直线检测——霍夫线变换
    x=min(x, y)
    x=min(x, y)
    算法 Tricks(三)—— 数组(序列)任意区间最小(大)值
    算法 Tricks(三)—— 数组(序列)任意区间最小(大)值
    分治法求解切割篱笆
    分治法求解切割篱笆
    GMM的EM算法实现
    秒杀多线程第二篇 多线程第一次亲热接触 CreateThread与_beginthreadex本质差别
  • 原文地址:https://www.cnblogs.com/13224ACMer/p/4409493.html
Copyright © 2011-2022 走看看