Hdoj 4939 stupid tower defense

It is obviously better to put red behind. DP [I] [J] indicates that there are j blue in the first I tower, and how many red are listed at the end.

Stupid tower defense Time Limit: 12000/6000 MS (Java/others) memory limit: 131072/131072 K (Java/Others)
Total submission (s): 599 accepted submission (s): 163


Problem descriptionfsf 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 shoshould have got M * y damage per second. the same, if you are already pass through K blue towers, the enemy shoshould have took T + K * z seconds every unit length.

FSF now wants to know the maximum damage the enemy can get.
Inputthere 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)
 
Outputfor each case, You shoshould 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: 12HintFor 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. 

 
Authoruestc
Source2014 multi-university training contest 7

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long int LL;LL n,x,y,z,t;LL dp[1600][1600];int main(){    int T_T,cas=1;    scanf("%d",&T_T);    while(T_T--)    {        cin>>n>>x>>y>>z>>t;        LL ans=0;        memset(dp,0,sizeof(dp));        for(LL i=2;i<=n;i++)        {            for(LL blue=0;blue<i;blue++)            {                if(blue)                {                    dp[i][blue]=max(dp[i-1][blue]+(t+blue*z)*(i-blue-1)*y                                ,dp[i-1][blue-1]+(t+(blue-1)*z)*(i-blue)*y);                }                else                {                    dp[i][0]=dp[i-1][0]+t*(i-1)*y;                }            }        }        for(LL red=0;red<=n;red++)        {            for(LL blue=0;blue<=n-red;blue++)            {                LL green=n-red-blue;                ans=max(ans,red*(x+green*y)*(blue*z+t)+dp[n-red][blue]);            }        }        printf("Case #%d: %I64d\n",cas++,ans);    }    return 0;}