hdu Stupid Tower Defense 4939 dp

Stupid Tower Defense

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 885    Accepted Submission(s): 264

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. 
   
  
  
  
  

题意：三种塔，红色在当前格子受到每单位时间x伤害，绿色在下一个位置每个单位时间受到y伤害，

蓝色在下一个位置每次停留时间增加z

可知红色一定是放在最后的，而且是连续放置，

dp[i][j]表示i个位置放置了j个绿色的塔，造成的最大伤害，

转移为dp[i][j] --->dp[i+1][j],dp[i+1][j+1]表示当前位置放置蓝色或者绿色塔，在下一时刻，造成的最大伤害

因为蓝色和绿色塔都是在下一个时刻起作用的。

然后枚举从第几个位置开始放红色的塔

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define max(x,y) x>y?x:y
#define maxn 1507
#define ll long long
ll dp[maxn][maxn];
ll n,dg,db,dr,st;
void init(int tt){
    ll t1,t2,ans=0;
    memset(dp,0,sizeof(dp));
    for(int i = 1;i < n; i++){
        for(int j = 0;j < i;j++){
            t1 = st+db*(i-j-1);
            t2 = st+db*(i-j);
            dp[i+1][j] = max(dp[i+1][j],dp[i][j]+j*dg*t2);
            dp[i+1][j+1] = dp[i][j]+(j+1)*dg*t1;
        }
    }
    for(int i = 1;i <= n; i++){
        for(int j = 0;j < i;j++){
            t1 = (i-j-1)*db+st;
            t2 = t1*(n-i+1);
            ans = max(ans,dp[i][j]+dr*t2+dg*(t2-t1)*j);
        }
    }
    printf("Case #%d: %I64d\n",tt,ans);
}
int main(){
    int t;
    scanf("%d",&t);
    for(int tt = 1; tt <= t; tt++){
        scanf("%I64d%I64d%I64d%I64d%I64d",&n,&dr,&dg,&db,&st);
        init(tt);
    }
    return 0;
}