HDOJ题目4939 Stupid Tower Defense（dp）

Stupid Tower Defense

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1759    Accepted Submission(s): 498

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

2014 Multi-University Training Contest 7

 

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看人家的http://blog.youkuaiyun.com/LYHVOYAGE/article/details/38531575

ac代码

#include<stdio.h>
#include<string.h>
#define max(a,b) (a>b?a:b)
__int64 dp[1510][1510];
__int64 n,x,y,z,t;
int main()
{
	int cas,cot=0;
	scanf("%d",&cas);
	while(cas--)
	{
		scanf("%I64d%I64d%I64d%I64d%I64d",&n,&x,&y,&z,&t);
		memset(dp,0,sizeof(dp));
		__int64 ans=n*t*x;
		int i,j;
		for(i=1;i<=n;i++)
		{
			for(j=0;j<=i;j++)
			{
				if(j==0)
					dp[i][j]=dp[i-1][j]+(i-1-j)*y*t;
				else
				{
					__int64 temp1,temp2;
					temp1=dp[i-1][j-1]+(i-j)*y*(t+(j-1)*z);
					temp2=dp[i-1][j]+(i-j-1)*y*(t+j*z);
					dp[i][j]=max(temp1,temp2);
				}
				ans=max(ans,dp[i][j]+(n-i)*x*(t+j*z)+(n-i)*y*(i-j)*(t+j*z));
			}
		}
		printf("Case #%d: %I64d\n",++cot,ans);
	}
}