DP大乱炖 A

最新推荐文章于 2025-08-17 18:08:09 发布

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原文链接：http://www.cnblogs.com/mybing/p/7358364.html

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There are two kinds of tasks, namely A and B. There are N workers and the i-th worker would like to finish one task A in ai minutes, one task B in bi minutes. Now you have X task A and Y task B, you want to assign each worker some tasks and finish all the tasks as soon as possible. You should note that the workers are working simultaneously.

input：In the first line there is an integer T(T<=50), indicates the number of test cases.
In each case, the first line contains three integers N(1<=N<=50), X,Y(1<=X,Y<=200). Then there are N lines, each line contain two integers ai, bi (1<=ai, bi <=1000).

output：For each test case, output “Case d: “ at first line where d is the case number counted from one, then output the shortest time to finish all the tasks.

Sample input：

Sample output：

Case 1: 2
Case 2: 4
Case 3: 6
题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=3433

题目分析：
你要明白最多所需花费时间为所有人中花费时间最多的人的时间，对于这个时间，我们可以二分得到，然后用DP来验证。
详细DP验证过程，请看代码注释

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<iomanip>
#include<algorithm>
using namespace std;
int t;
int n,m,f[55][510],a[51],b[51],x,y,l,r;
bool work(int p)
{
    memset(f,-1,sizeof(f));
    //每次验证之前需要把数组全部赋值为负 
    f[0][0]=0;
    for(int i=1;i<=n;i++)//枚举每一个工人 
    {
        for(int j=0;j<=x;j++)//在此处，我们做一个类似于背包问题的dp，把所需要做A时间看为背包代价 
        {
            f[i][j]=f[i-1][j];
            for(int k=0;k<=j;k++)//在做j个A事件时能做的最多的B事件 
            {
                if(p-k*a[i]<0) break;
                if(f[i-1][j-k]!=-1)
                f[i][j]=max(f[i][j],f[i-1][j-k]+(p-k*a[i])/b[i]);
            }
        }
    }
    if(f[n][x]>=y) return true;//如果在做完所有A事件的时候所能做最多数量的B事件。 
    else return false;
}
int main()
{
    scanf("%d",&t);
    for(int qw=1;qw<=t;qw++)
    {
        scanf("%d%d%d",&n,&x,&y);
        for(int i=1;i<=n;i++){scanf("%d%d",&a[i],&b[i]);}
        l=1,r=1000000;
        while(l+1<r)//二分枚举最短时间 
        {
            int mid=(l+r)/2;
            if(work(mid))r=mid;
            else l=mid;
        }
        cout<<"Case "<<qw<<": ";
        if(work(l))cout<<l<<endl;
        else cout<<r<<endl;
    }
    return 0;
}