hdoj4939Stupid Tower Defense【dp】

Stupid Tower Defense

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2055    Accepted Submission(s): 558

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

题意：一个怪物在一条直线上走一个单位需要t秒时间总共有n个单位长度；在每单位长度可以选择放三种塔红塔对过该位置的怪物每秒造成x点伤害。绿塔在怪物经过该位置以后的每个位置对该怪物造成每秒y的伤害。蓝塔对经过该位置的怪物以后每经过一个位置的时间增加z

设dp[i][j]表示在前i个位置放j个蓝塔对怪物造成的最大伤害。对i-j的位置放lv塔对剩下的n-i个位置放红塔；

所以有动态转移方程放绿塔dp[i][j]=max(dp[i][j],dp[i-1][j]+(i-j-1)*y*(t+j*z));放蓝塔：dp[i][j]=max(dp[i][j],dp[i-1][j-1]+(i-j)*y*(t+(j-1)*z));

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<list>
#include<queue>
using namespace std;
const int maxn=1510;
long long dp[maxn][maxn];
int main()
{
	int T,test=1;
	long long n,x,y,z,t,i,j,ans;
	scanf("%d",&T);
	while(T--){
		scanf("%lld%lld%lld%lld%lld",&n,&x,&y,&z,&t);
		memset(dp,0,sizeof(dp));
		ans=n*x*t;
		for(i=1;i<=n;++i){
			for(j=0;j<=i;++j){
				dp[i][j]=max(dp[i][j],dp[i-1][j]+(i-j-1)*y*(t+j*z));
				if(j){
					dp[i][j]=max(dp[i][j],dp[i-1][j-1]+(i-j)*y*(t+(j-1)*z));
				}
				ans=max(ans,dp[i][j]+(n-i)*(x+(i-j)*y)*(t+j*z)); 
			}
		}
		printf("Case #%d: %lld\n",test++,ans);
	}
	return 0;
}