dp（2014多校联合）

最新推荐文章于 2018-07-22 13:03:24 发布

原创最新推荐文章于 2018-07-22 13:03:24 发布 · 461 阅读

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本文介绍了一款塔防游戏中的策略问题，通过动态规划的方法来计算敌人可能遭受的最大伤害。文章详细阐述了游戏地图的设定及不同类型的塔楼对敌人的伤害计算方式，并给出了一段C++代码实现。

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Stupid Tower Defense

Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 256 Accepted Submission(s): 56

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.

dp[i][j]表示前i个单位有j个蓝塔，那么转移方程是：dp[i][j]=max(dp[i-1][j-1]+(LL)((LL)i-j)*y*(t+(j-1)*z),dp[i-1][j]+(i-1-j)*y*(t+j*z))

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
using namespace std;
const int maxn=1600;
const long long INF=1e16;
typedef long long LL;
LL dp[maxn][maxn];
LL n,x,y,z,t;
void solve()
{
    LL ans=-INF;
    memset(dp,0,sizeof(dp));
    for(int i=1;i<=n;i++)
    {
        for(int j=0;j<=i;j++)
        {
            dp[i][j]=max(dp[i][j],dp[i-1][j]+(i-1-j)*y*(t+j*z));
            if(j>=1)dp[i][j]=max(dp[i][j],dp[i-1][j-1]+(LL)((LL)i-j)*y*(t+(j-1)*z));
            ans=max(ans,dp[i][j]+x*(n-i)*(t+j*z)+(n-i)*(i-j)*y*(t+z*j));
        }
    }
    ans=max(ans,n*x*t);//全是红塔的时候
    cout<<ans<<endl;
}
int main()
{
    int T,cas=1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%I64d%I64d%I64d%I64d%I64d",&n,&x,&y,&z,&t);
        printf("Case #%d: ",cas++);
        solve();
    }
    return 0;
}