HDU 4939 Stupid Tower Defense

Stupid Tower Defense

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 860    Accepted Submission(s): 257

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.
 

 

官方的解释；

按官方的解释的代码：

#include <cstdio>#include <cmath>#include <cstdlib>#include <cstring>#include <vector>#include <map>#include <iostream>#include <algorithm>using namespace std;#define LL __int64#define N 1605LL dp[N][N];LL n,x,y,z,t;int main() {    int T;    cin>>T;    for(int cas=1; cas<=T; cas++) {        scanf("%I64d %I64d %I64d %I64d %I64d",&n,&x,&y,&z,&t);        printf("Case #%d: ",cas);        for(int i=0;i<N;i++){            for(int j=0;j<N;j++){                dp[i][j]=0;            }        }        LL ans=n*t*x;        for(LL i=1;i<=n;i++){            for(LL j=0;j<=i;j++){                if(j==0){                    dp[i][j]=dp[i-1][j]+t*(i-j-1)*y;                }                else{                    LL temp1=dp[i-1][j-1]+(i-j)*y*(t+(j-1)*z);//第j个位置放蓝塔                    LL temp2=dp[i-1][j]+(i-j-1)*y*(t+j*z);//第j个位置不放蓝塔                    dp[i][j]=max(temp1,temp2);                }                ans=max(ans,dp[i][j]+(n-i)*x*(t+j*z)+(n-i)*(t+j*z)*(i-j)*y);            }        }        printf("%I64d\n",ans);    }    return 0;}