HDU 4939 一个塔防游戏

最新推荐文章于 2017-08-16 19:45:39 发布

hitwhacmer1

最新推荐文章于 2017-08-16 19:45:39 发布

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本文探讨了一款塔防游戏中的最优策略，通过动态规划方法计算敌方单位可能遭受的最大伤害。考虑不同塔的功能特性，如减速塔、即时伤害塔及延迟伤害塔，建立数学模型并给出实现代码。

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Stupid Tower Defense
Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 601 Accepted Submission(s): 164

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.

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.

今天跟着学长做今年的多校，被虐的死死的，自己有思路的只有hdu4941跟这一道dp的题，哎，还是太弱了，做的时候想到了要用点贪心把red放最后枚举个数，然后对蓝塔跟绿塔进行dp；

dp[i][j]表示前i个塔中有j个蓝塔，剩余的全都弄成红塔

dp[i][j]=max((dp[i-1][j]+(i-j-1))*y*(j*z+t)),(dp[i-1][j-1]+(i-j)*y*((j-1)*z+t)))

这个就是状态转移方程。

然后再加上n-i个red塔的伤害：dp[i][j]+(n-i)*x*(t+j*z)+(n-i)*(t+j*z)*(i-j)*y，然后这些里面找最大的。

状态转移方程我在做的时候都写出来了，然后最后真正写的时候一些小的情况处理不好。

还是dp太弱，做的dp题太少了，想起来一个队友说的话，我们弱校之所以弱最重要的就是dp意识太弱了，或许很有道理

哎，不发牢骚了，发代码：

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
long long dp[2000][2000];
long long n,x,y,z,t,Max,i,j;
int main()
{
    int T,tt;
    scanf("%d",&T);
    for(tt=1; tt<=T; tt++)
    {
        scanf("%I64d%I64d%I64d%I64d%I64d",&n,&x,&y,&z,&t);
        memset(dp,0,sizeof(dp));
        Max=n*x*t;
        for(i=1; i<=n; i++)
            for(j=0; j<=i; j++)
            {
                if(j==0)
                    dp[i][j]=dp[i-1][j]+t*(i-j-1)*y;
                else
                    dp[i][j]=max((dp[i-1][j]+(max(0LL,i-j-1))*y*(j*z+t)),(dp[i-1][j-1]+(i-j)*y*((j-1)*z+t)));
                Max=max(dp[i][j]+(n-i)*x*(t+j*z)+(n-i)*(t+j*z)*(i-j)*y,Max);
            }
       printf("Case #%d: %I64d\n",tt,MAX);
    }
    return 0;
}