/*Stupid Tower DefenseTime Limit: 12000/6000 MS (java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 2187 Accepted Submission(s): 603PRoblem 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 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.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 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 Input12 4 3 2 1Sample OutputCase #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.//关键在于确定状态,通过最后一个格子区分状态,确保涵盖所有状态*/#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn = 1507;typedef long long LL;LL Dp[maxn][maxn];int main(){ int T; cin >> T; for (int k = 1; k <= T; k++) { LL n, x, y, z, t; cin >> n >> x >> y >> z >> t; LL ans = (x*n*t); memset(Dp, 0, sizeof(Dp)); for (int i = 1; i <= n; i++) { for (int j = 0; j <= i; j++) { if (j == 0) { Dp[i][j] = Dp[i - 1][j] + (i - 1)*y*t; } else {//从对于一个n个格子的情况可以分为两种情况,一种是最后一个格子是绿的,一种是最后一个格子是蓝的 Dp[i][j] = max(Dp[i - 1][j] + (t + z*j)*y*max((i - j - 1), 0), Dp[i - 1][j - 1] + ((j - 1)*z + t)*y*(i - j)); } ans = max(ans, (long long)(Dp[i][j] + (n - i)*(x + y*(i - j))*(t + j*z))); } } printf("Case #%d: %I64d/n", k, ans); } return 0;}
