red放到后面显然更优，dp【i】【j】表示前i个塔里有j个blue，最后枚举有多少个red
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 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
HintFor 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
#include
#include
#include
#include
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