zoukankan      html  css  js  c++  java
  • ZOJ 3703 Happy Programming Contest(0-1背包)

    http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3703


    Happy Programming Contest

    Time Limit: 2 Seconds      Memory Limit: 65536 KB

    In Zhejiang University Programming Contest, a team is called "couple team" if it consists of only two students loving each other. In the contest, the team will get a lovely balloon with unique color for each problem they solved. Since the girl would prefer pink balloon rather than black balloon, each color is assigned a value to measure its attractiveness. Usually, the boy is good at programming while the girl is charming. The boy wishes to solve problems as many as possible. However, the girl cares more about the lovely balloons. Of course, the boy's primary goal is to make the girl happy rather than win a prize in the contest.

    Suppose for each problem, the boy already knows how much time he needs to solve it. Please help him make a plan to solve these problems in strategic order so that he can maximize the total attractiveness value of balloons they get before the contest ends. Under this condition, he wants to solve problems as many as possible. If there are many ways to achieve this goal, he needs to minimize the total penalty time. The penalty time of a problem is equal to the submission time of the correct solution. We assume that the boy is so clever that he always submit the correct solution.

    Input

    The first line of input is an integer N (N < 50) indicating the number of test cases. For each case, first there is a line containing 2 integers T (T <= 1000) and n (n <= 50) indicating the contest length and the number of problems. The next line contains n integers and the i-th integer ti (ti <= 1000) represents the time needed to solve the ith problem. Finally, there is another line containing n integers and the i-th integer vi (vi <= 1000) represents the attractiveness value of the i-th problem. Time is measured in minutes.

    Output

    For each case, output a single line containing 3 integers in this order: the total attractiveness value, the number of problems solved, the total penalty time. The 3 integers should be separated by a space.

    Sample Input

    2
    300 10
    10 10 10 10 10 10 10 10 10 10
    1 2 3 4 5 6 7 8 9 10
    300 10
    301 301 301 301 301 301 301 301 301 301
    1000 1000 1000 1000 1000 1000 1000 1000 1000 1000
    
    

    Sample Output

    55 10 550
    0 0 0
    


    思路:0-1背包


    <span style="font-size:18px;">#include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    const int maxn=1010;
    int value[maxn],a[maxn],fv[maxn];
    int vis[55][maxn];
    int order[55];
    int main()
    {
        int N,length,n;
        cin>>N;
        while(N--)
        {
            int time=0,sumtime=0,numbers=0;
            memset(vis,0,sizeof(vis));
            memset(fv,0,sizeof(fv));
            cin>>length>>n;
            for(int i=1;i<=n;i++)cin>>a[i];
            for(int i=1;i<=n;i++)cin>>value[i];
            for(int i=1;i<=n;i++)
            {
               for(int j=length;j>=a[i];j--)
               {
                   if(fv[j]<fv[j-a[i]]+value[i])
                   {
                       fv[j]=fv[j-a[i]]+value[i];
                       vis[i][j]=1;
                   }
               }
            }
            int cnt=0;
            int i=n;
            int j=length;
            for(;i>0;i--)
            {
                if(vis[i][j]==1)
                {
                    order[cnt++]=a[i];
                    j -= a[i];
                }
            }
            sort(order,order+cnt);
            for(int i=0;i<cnt;i++)
            {
                time +=order[i];
                sumtime +=time;
            }
            cout<<fv[length]<< " " << cnt << " " << sumtime << endl;
        }
        return 0;
    }</span>



  • 相关阅读:
    灾难 BZOJ 2815
    消耗战 BZOJ 2286
    征途 BZOJ 4518
    纸箱堆叠 BZOJ 2253
    Gate Of Babylon BZOJ 1272
    std::string::npos mean
    [转]整理索引碎片,提升SQL Server速度
    笔记本win7制作wifi
    关闭linux下的使用的端口
    linux多线程
  • 原文地址:https://www.cnblogs.com/slgkaifa/p/7275689.html
Copyright © 2011-2022 走看看