dp --- hdu 4939 : Stupid Tower Defense

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.

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
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 long
using 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;
}