Stupid Tower Defense
Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 151 Accepted Submission(s): 32
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.
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)
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
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
思路:红塔放在最后,绿塔和蓝塔dp
注意:绿塔和蓝塔的效果都是延迟的,当前塔没有效果,要到下一格。
1 #include<iostream> 2 #include<cstring> 3 #include<cstdlib> 4 #include<cstdio> 5 #include<algorithm> 6 #include<cmath> 7 #include<queue> 8 9 #define N 1505 10 #define M 105 11 #define mod 1000000007 12 #define mod2 100000000 13 #define ll long long 14 #define maxi(a,b) (a)>(b)? (a) : (b) 15 #define mini(a,b) (a)<(b)? (a) : (b) 16 17 using namespace std; 18 19 ll i,j; 20 ll n; 21 int T; 22 ll x,y,z,t; 23 ll ans; 24 ll te; 25 ll k,m,p; 26 ll dp[N][N]; 27 28 int main() 29 { 30 freopen("data.in","r",stdin); 31 scanf("%d",&T); 32 for(int cnt=1;cnt<=T;cnt++) 33 { 34 35 memset(dp,0,sizeof(dp)); 36 scanf("%I64d%I64d%I64d%I64d%I64d",&n,&x,&y,&z,&t); 37 ans=0; 38 for(i=0;i<=n;i++){ 39 for(k=0;k<=i;k++){ 40 m=i-k;p=n-m-k; 41 if(k==0){ 42 if(m==0) ans=max(ans,t*n*x); 43 else{ 44 dp[m][k]=dp[m-1][k]+(t+k*z)*(m-1)*y; 45 ans=max(ans,dp[m][k]+p*(t+k*z)*(x+m*y)); 46 } 47 } 48 else{ 49 if(m==0){ 50 dp[m][k]=dp[m][k-1]+(t+(k-1)*z)*m*y; 51 ans=max(ans,dp[m][k]+p*(t+k*z)*(x+m*y)); 52 } 53 else{ 54 dp[m][k]=max(dp[m-1][k]+(t+k*z)*(m-1)*y,dp[m][k-1]+(t+(k-1)*z)*m*y); 55 ans=max(ans,dp[m][k]+p*(t+k*z)*(x+m*y)); 56 } 57 } 58 59 } 60 } 61 printf("Case #%d: %I64d ",cnt,ans); 62 } 63 return 0; 64 }