PAT顶级 1002. Business (35)
As the manager of your company, you have to carefully consider, for each project, the time taken to finish it, the deadline, and the profit you can gain, in order to decide if your group should take this project. For example, given 3 projects as the following:
Project[1] takes 3 days, it must be finished in 3 days in order to gain 6 units of profit.
Project[2] takes 2 days, it must be finished in 2 days in order to gain 3 units of profit.
Project[3] takes 1 day only, it must be finished in 3 days in order to gain 4 units of profit.
You may take Project[1] to gain 6 units of profit. But if you take Project[2] first, then you will have 1 day left to complete Project[3] just in time, and hence gain 7 units of profit in total. Notice that once you decide to work on a project, you have to do it from beginning to the end without any interruption.
Input Specification
Each input file contains one test case. For each case, the first line gives a positive integer N(<=50), and then followed by N lines of projects, each contains three numbers P, L, and D where P is the profit, L the lasting days of the project, and D the deadline. It is guaranteed that L is never more than D, and all the numbers are non-negative integers.
Output Specification
For each test case, output in a line the maximum profit you can gain.
Sample Input
4
7 1 3
10 2 3
6 1 2
5 1 1
Sample Output
18
设dp[i][j]表示考虑到第i个项目，前一个项目结束时间为j的最大收益,那么可得DP方程为:
dp[i][j+a[i].l]=max(dp[i][j+a[i].l],dp[k][j]) if(j+a[i].l<=a[i].d)
由于没有说明数据范围，只能使用map表示二维dp数组了
#include <map>
#include <iostream>
#include <algorithm>
using namespace std;
struct node
{
int p,l,d;
node() {}
node(int tp,int tl,int td)
{
p=tp;
l=tl;
d=td;
}
bool operator < (const node &an) const
{
return d<an.d;
}
} a[55];
typedef long long ll;
map<ll,map<ll,ll> >dp;
int main()
{
int n,p,l,d;
while(cin>>n)
{
for(int i=1; i<=n; i++)
{
cin>>p>>l>>d;
a[i]=node(p,l,d);
}
a[0]=node(0,0,0);
sort(a+1,a+1+n);
dp.clear();
ll ans=0;
for(int i=1; i<=n; i++)
{
for(int j=0; j<=a[i-1].d; j++)
{
if(j+a[i].l<=a[i].d)
{
for(int k=0;k<i;k++){
dp[i][j+a[i].l]=max(dp[i][j+a[i].l],dp[k][j]+a[i].p);
ans=max(ans,dp[i][j+a[i].l]);
}
}
}
}
cout<<ans<<endl;
}
return 0;
}