HDU4939 dp

http://acm.hdu.edu.cn/showproblem.php?pid=4939

Problem Description

FSF is addicted to a stupid tower defense game. The goal of tower defense games is to try to stop enemies from crossing a map by building traps to slow them down and towers which shoot at them as they pass.

The map is a line, which has n unit length. We can build only one tower on each unit length. The enemy takes t seconds on each unit length. And there are 3 kinds of tower in this game: The red tower, the green tower and the blue tower. 

The red tower damage on the enemy x points per second when he passes through the tower.

The green tower damage on the enemy y points per second after he passes through the tower.

The blue tower let the enemy go slower than before (that is, the enemy takes more z second to pass an unit length, also, after he passes through the tower.)

Of course, if you are already pass through m green towers, you should have got m*y damage per second. The same, if you are already pass through k blue towers, the enemy should have took t + k*z seconds every unit length.

FSF now wants to know the maximum damage the enemy can get.

 

Input

There are multiply test cases.

The first line contains an integer T (T<=100), indicates the number of cases. 

Each test only contain 5 integers n, x, y, z, t (2<=n<=1500,0<=x, y, z<=60000,1<=t<=3)

 

Output

For each case, you should output "Case #C: " first, where C indicates the case number and counts from 1. Then output the answer. For each test only one line which have one integer, the answer to this question.

 

Sample Input

  

   1
2 4 3 2 1
  

 

Sample Output

  

   Case #1: 12


   

    

     Hint
    
For the first sample, the first tower is blue tower, and the second is red tower. So, the total damage is 4*(1+2)=12 damage points.
   
  

官方解题思路：

       

/**
hdu 4939  dp
*/
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;

LL dp[1505][1505];
LL n,x,y,z,t;

int main()
{
    int T,tt=1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%I64d%I64d%I64d%I64d%I64d",&n,&x,&y,&z,&t);
        memset(dp,0,sizeof(dp));
        LL ans=n*t*x;
        for(int i=1;i<=n;i++)
        {
            for(int j=0;j<=i;j++)
            {
                if(j==0)
                    dp[i][j]=dp[i-1][j]+t*(i-1)*y;
                else
                    dp[i][j]=max(dp[i-1][j]+(j*z+t)*max(0,i-1-j)*y,dp[i-1][j-1]+((j-1)*z+t)*(i-j)*y);
                ans=max(ans,dp[i][j]+(n-i)*(x+(i-j)*y)*(t+j*z));
            }
        }
        printf("Case #%d: %I64d\n",tt++,ans);
    }
    return 0;
}