hdu 4939 Stupid Tower Defense dp

最新推荐文章于 2017-08-16 19:45:39 发布

Excelsior_kereo

最新推荐文章于 2017-08-16 19:45:39 发布

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分类专栏：简单dp ------------dp------------- 文章标签： dp

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本文探讨了一个简化版的塔防游戏StupidTowerDefense，分析了如何通过合理布局不同类型的塔来最大化对敌人的伤害。游戏地图为一条直线，包含红塔、绿塔和蓝塔三种塔型，每种塔都有特定的效果。文章通过动态规划的方法求解最优解。

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Stupid Tower Defense

Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 497 Accepted Submission(s): 130

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

思路：简单dp。设dp[ j ][ i - j ]为前i个位置有 j 个绿塔，i - j 个蓝塔，当敌人走到 i 时受到的最大伤害。那么我们很容易推出 dp[ j ] [ i - j ] 是由dp[ j - 1 ][ i - j ] 和 dp [ j ] [ i - j -1 ]

转移而来。dp[ j ][ i - j ]=max(dp[ j ][ i- j ],dp[ j - 1 ][ i - j ]+( j - 1 ) * y * ( t + ( i - j ) * z ) ); dp[ j ] [ i - j ] = max ( dp[ j ] [ i - j ] , dp[ j ][ i - j - 1 ] +j * y * ( t + ( i - j - 1 ) * z ) ); 同时维护ans,

ans=max(ans,dp[ j ] [ i - j ] + ( n - i ) * ( t + ( i - j ) * z ) * x + j * y * ( n - i ) * ( t + ( i - j ) * z ) ) ; 即假设走完i之后后面只有红塔了，累加个红塔的攻击力以及绿塔、蓝塔之后的攻击力，详见代码：

// file name: 1.cpp //
// author: kereo //
// create time:  2014年08月12日 星期二 17时52分23秒 //
//***********************************//
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<set>
#include<map>
#include<vector>
#include<stack>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long ll;
const int MAXN=2000+100;
#define L(x) (x<<1)
#define R(x) (x<<1|1)
ll n,x,y,z,t;
ll dp[MAXN][MAXN];
void init()
{
    memset(dp,0,sizeof(dp));
}
int main()
{
    //freopen("text","r",stdin);
    int T,kase=0;
    scanf("%d",&T);
    while(T--){
        printf("Case #%d: ",++kase);
        init();
        scanf("%I64d%I64d%I64d%I64d%I64d",&n,&x,&y,&z,&t);
        ll ans=n*x*t;
        for(ll i=1;i<n;i++){
            for(ll j=0;j<=i;j++){
                if(j)
                    dp[j][i-j]=max(dp[j][i-j],dp[j-1][i-j]+(j-1)*y*(t+(i-j)*z));
                if(i-j)
                    dp[j][i-j]=max(dp[j][i-j],dp[j][i-j-1]+j*y*(t+(i-j-1)*z));
                ans=max(ans,dp[j][i-j]+(n-i)*(t+(i-j)*z)*x+j*y*(n-i)*(t+(i-j)*z));
            }
        }
        printf("%I64d\n",ans);
    }
    return 0;
}