dp --- hdu 4939 : Stupid Tower Defense
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Stupid Tower Defense
Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1219 Accepted Submission(s): 361
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
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
Mean:
经典的塔防类游戏。
敌人要通过一条过道,你有三种塔:红塔---敌人经过该塔时每秒受到x点伤害; 绿塔---敌人经过该塔后,每秒受到y点伤害; 蓝塔---敌人经过该塔后,经过每座塔的时间变慢z秒。现在要你安排这三种塔,使得对敌人的伤害最大。
analyse:
分析可知,红塔要放到后面。
然后我们枚举红塔的数量i,对前n-i座塔进行dp。
dp[i][j]----表示前i座塔中,放j座蓝塔和i-j座绿塔所造成的最大伤害。
x y z t
状态转移方程:
dp[i][j]=max(dp[i-1][j-1]+y*(i-j)*(t+(j-1)*z),dp[i-1][j]+(i-1-j)*y*(t+j*z))
。
Time complexity:O(n^2)
Source code:
//Memory Time// 18424K 1796MS// by : Snarl_jsb#include<algorithm>#include<cstdio>#include<cstring>#include<cstdlib>#include<iostream>#include<vector>#include<queue>#include<stack>#include<string>#include<climits>#include<cmath>#define MAX 1520#define LL long longusing namespace std;LL dp[MAX][MAX];int main(){ LL T,kase=1; cin>>T; while(T--) { LL n,x,y,z,t,damage; scanf("%I64d %I64d %I64d %I64d %I64d",&n,&x,&y,&z,&t); printf("Case #%I64d: ",kase++); memset(dp,0,sizeof(dp)); dp[1][1]=x; LL ans=n*x*t; //全部放红塔的伤害值 for(LL i=1;i<=n;i++) //枚举前i个单位长度 { for(LL j=0;j<=i;j++) // 枚举前i个单位中蓝塔的数量j { if(j==0) dp[i][j]=dp[i-1][j]+y*(i-1)*t; else { LL tmp1=dp[i-1][j-1]+y*(i-j)*(t+z*(j-1)); // 第j座放蓝塔 LL tmp2=dp[i-1][j]+y*(i-1-j)*(t+z*j); // 第j座放绿塔 dp[i][j]=max(tmp1,tmp2); } damage=dp[i][j]+(n-i)*x*(t+z*j)+(n-i)*(i-j)*y*(t+z*j); ans=max(ans,damage); } } cout<<ans<<endl; } return 0;}
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