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

12 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;}