• ELEVATOR 南邮NOJ 1996


    ELEVATOR

    时间限制(普通/Java) : 1000 MS/ 3000 MS          运行内存限制 : 65536 KByte
    总提交 : 93            测试通过 : 24 

    题目描述

    In the library of NUPT there is an elevator. The elevator can lift no more than W kg.
    And you know in NUPT the library is always crowded.
    Here are N people want to get into the elevator. Each people have two facts, the weight and IQ.
    I want you to find the minimum of the combined IQ of the people outside the elevator when the combined weight of the people in the elevator is no more than W. 
    For example: when W,n=200,4
            The weight of people are: 65 50 85 55
            The IQ of peole are:     46 18 70 21
    65+50+85+55=255>W, so the elevator can lift them all.
    We chose to lift the 1st,2nd,3rd people, the sum of weight is 200<=W and the sum of  IQ of people out of the elevator is 21.
    So you have to output 21.

    输入

    In the first line there is an integer T indicating there T test cases.
    In the first of each case, there are two integers W,n.W is the max weight of peole in the elevator, and there are n people.
    In the next line there are n integers wi. The ith integer is the ith people’s weight.
    In the next line there are n integers vi. The ith integer is the ith people’s IQ.

    输出

    For each case you only output one integer. The combined IQ of the people outside the elevator.

    样例输入

    2
    0 1
    1
    1
    200 4
    65 50 85 55
    46 18 70 21

    样例输出

    1
    21

    提示

    null

    #include<iostream>
    #include<cstring>
    #include<cstdlib>
    #include<cstdio>
    using namespace std;
    int main()
    {
        int dp[101][1001],m,T,C,w[101],val[101],i,j;
         scanf("%d",&C);
         while(C--){
         int sum=0;  //累加器
         scanf("%d%d",&T,&m);
         for(i=1;i<=m;i++)
             scanf("%d",&w[i]);  //体重
         for(i=1;i<=m;i++)
         {
             scanf("%d",&val[i]); //IQ值
             sum+=val[i];
         }
        memset(dp,0,sizeof(dp));  //置零
        for(i=1;i<=m;i++)
        for(j=0;j<=T;j++)
        {
            if(j>=w[i])
                dp[i][j]=max(dp[i-1][j],dp[i-1][j-w[i]]+val[i]);//放还是不放的选择
            else
                dp[i][j]=dp[i-1][j];
        }
        printf("%d
    ",sum-dp[m][T]);
        }
        return 0;
    }      //01背包问题的另一种形式

    版权声明:本文为博主原创文章,未经博主允许不得转载。

  • 相关阅读:
    OpenResty 作者章亦春访谈实录
    linux 下C语言学习路线
    swift 集合类型
    不懂技术的人不要对懂技术的人说这很容易实现
    java学习笔记7--抽象类与抽象方法
    java学习笔记6--类的继承、Object类
    python练习题
    python的里字典和列表
    python里list列表,tuple元组内部功能介绍
    python里float和long内部功能及字符串str介绍
  • 原文地址:https://www.cnblogs.com/Tobyuyu/p/4965591.html
Copyright © 2020-2023  润新知