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HDU 4939 Stupid Tower Defense(dp+贪心)

2024-07-17 12:46:00   229人阅读

dp[i][j]表示到了第i步放了j个减速，造成的伤害。我们用贪心的策略把造成一段伤害的放在最后面，造成持续伤害的与减速放在前i个中这样得到的伤害是最高的。

所以前(i,j)中的伤害为dp[i][j] = max(dp[i-1][j]+(j*z+t)*(max(0LL, i-1-j))*y, dp[i-1][j-1]+((j-1)*z+t)*(i-j)*y);

每次造成的伤害就为：dp[i][j]+(n-i)*(j*z+t)*(x+(i-j)*y)。然后取一个最大值，就是伤害的最大值了。

Stupid Tower Defense

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1194    Accepted Submission(s): 350


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



#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <ctime>
#include <map>
#include <set>
#define eps 1e-9
///#define M 1000100
#define LL __int64
///#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)

using namespace std;

const int maxn = 2010;

LL dp[maxn][maxn];

int main()
{
    int T;
    cin >>T;
    int Case = 1;
    while(T--)
    {
        LL n, x, y, z, t;
        cin >>n>>x>>y>>z>>t;
        memset(dp, 0, sizeof(dp));
        LL ans = n*t*x;
        for(LL i = 1; i <= n; i++)
        {
            for(LL j = 0; j <= i; j++)
            {
                if(j == 0) dp[i][j] = dp[i-1][j]+t*(i-1-j)*y;
                else dp[i][j] = max(dp[i-1][j]+(j*z+t)*(max(0LL, i-1-j))*y, dp[i-1][j-1]+((j-1)*z+t)*(i-j)*y);
                ans = max(ans, dp[i][j]+(n-i)*(j*z+t)*(x+(i-j)*y));
            }
        }
        cout<<"Case #"<<Case++<<": "<<ans<<endl;
    }
    return 0;
}



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