dp --- hdu 4939 : Stupid Tower DefenseStupid Tower Defense

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 1219    Accepted Submission(s): 361

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.

Author

UESTC

Source

<a href="http://www.cnblogs.com/search.php?field=problem&amp;key=2014%20Multi-University%20Training%20Contest%207&amp;source=1&amp;searchmode=source">2014 Multi-University Training Contest 7</a>

Mean: 

經典的塔防類遊戲。

敵人要通過一條過道，你有三種塔：紅塔---敵人經過該塔時每秒受到x點傷害；  綠塔---敵人經過該塔後，每秒受到y點傷害； 藍塔---敵人經過該塔後，經過每座塔的時間變慢z秒。現在要你安排這三種塔，使得對敵人的傷害最大。

analyse:

分析可知，紅塔要放到後面。

然後我們枚舉紅塔的數量i,對前n-i座塔進行dp。

dp[i][j]----表示前i座塔中，放j座藍塔和i-j座綠塔所造成的最大傷害。

x y z t

狀态轉移方程：

dp[i][j]=max(dp[i-1][j-1]+y*(i-j)*(t+(j-1)*z),dp[i-1][j]+(i-1-j)*y*(t+j*z))

。

Time complexity：O(n^2)

Source code：