hdu 4939 Stupid Tower Defense dp

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Stupid Tower Defense

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1986    Accepted Submission(s): 545


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.
 

Author
UESTC
 

Source
2014 Multi-University Training Contest 7
 

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#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>#include<climits>#include<queue>#include<vector>#include<map>#include<sstream>#include<set>#include<stack>#include<cctype>#include<utility>#pragma comment(linker, "/STACK:102400000,102400000")#define PI (4.0*atan(1.0))#define eps 1e-10#define sqr(x) ((x)*(x))#define FOR0(i,n)  for(int i=0 ;i<(n) ;i++)#define FOR1(i,n)  for(int i=1 ;i<=(n) ;i++)#define FORD(i,n)  for(int i=(n) ;i>=0 ;i--)#define  lson   ind<<1,le,mid#define rson    ind<<1|1,mid+1,ri#define MID   int mid=(le+ri)>>1#define zero(x)((x>0? x:-x)<1e-15)#define mk    make_pair#define _f     first#define _s     secondusing namespace std;//const int INF=    ;typedef long long ll;//const ll inf =1000000000000000;//1e15;//ifstream fin("input.txt");//ofstream fout("output.txt");//fin.close();//fout.close();//freopen("a.in","r",stdin);//freopen("a.out","w",stdout);const int INF =0x3f3f3f3f;const int maxn= 1500+20   ;//const int maxm=    ;ll n,x,y,z,t;ll maxi,dp[maxn][maxn];//dp[i][j]表示前面放了i个塔,其中有j个蓝塔。int main(){    int T,kase=0;scanf("%d",&T);    while(T--)    {        scanf("%lld%lld%lld%lld%lld",&n,&x,&y,&z,&t);        maxi=t*n*x;        //注意状态的表示,以及对于任意一种方法 红塔应该放在最后。        //但没想到红塔的伤害可以在O(1)时间内算出。        //就是枚举i和j然后可以求出红塔的伤害。这里用了“红塔应该放在最后”;        for(int i=1;i<=n;i++)        {            for(int j=0;j<=i;j++)            {                if(!j)   dp[i][j]=dp[i-1][j]+ t*(i-1)*y ;   //此步放绿塔                else  dp[i][j]=max(dp[i-1][j]  + (t+           j*z) *max(0,(i-j-1))*y,//绿                                   dp[i-1][j-1]+ (t+ max(0,j-1 )*z) *(i-j)*y     );//蓝                maxi=max(maxi, dp[i][j]+(t+j*z)* (n-i)*(x+ (i-j)*y    )  );//之后放红塔            }        }        printf("Case #%d: %lld\n",++kase,maxi);    }    return 0;}


 
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