hdu 4939 Stupid Tower Defense DP

Stupid Tower Defense

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1181    Accepted Submission(s): 346

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

重点在于把三个要素巧妙转换成2个！
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<string>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<list>
#include<map>
#include<set>
using namespace std;
int n,x,y,z,pertime;
typedef long long LL;
LL dp[1510][1510];

void work(){
    memset(dp, 0, sizeof(dp));
    LL redgoal,tp;
    LL ans =1LL * n * pertime * x; // 全部是red ;

    for(int i = 1 ; i <= n ; i ++){
        for(int bn = 0; bn <= i; bn ++){
            if(i - 1 >= bn){
                    // 第 i 个格子建的是 green 塔；
                dp[i][bn] = dp[i - 1][bn] + 1LL * (n - i) * (pertime + bn * z) * y;
                                // 这个塔的后面有 n - i 个格子，当前时间为 t+bn*z 分数为 y;
                redgoal = 1LL * (n - i) * x * (pertime + bn * z);
                ans = max(ans , dp[i][bn] + redgoal);
;
            }
            if(bn > 0){
                    //第i位置建立的是blue， 前面有 i - bn 个 green 在 第 i 个格子
                    // 之后（n - i ）的收益将增加 y * z *(n - i);
                tp = dp[i - 1][bn - 1] + 1LL * (i - bn) * (n - i) * z * y;
                redgoal = 1LL * (n - i) * x * (pertime + bn * z);
                ans = max(ans, tp + redgoal);
                dp[i][bn] = max(dp[i][bn], tp);
            }
        }
    }
    cout << ans << endl;
}
int main()
{

//	freopen("in.in","r",stdin);
    int T;
    scanf("%d",&T);
    for(int ca = 1; ca <= T; ca ++){
        scanf("%d%d%d%d%d",&n, &x, &y, &z, &pertime);
        printf("Case #%d: ",ca);
        work();
    }

	return 0;
}