E. FTL

E. FTL

Monocarp is playing a video game. In the game, he controls a spaceship and has to destroy an enemy spaceship.

Monocarp has two lasers installed on his spaceship. Both lasers $1$ and $2$ have two values:

$p_i$ — the power of the laser;
$t_i$ — the reload time of the laser.
When a laser is fully charged, Monocarp can either shoot it or wait for the other laser to charge and shoot both of them at the same time.

An enemy spaceship has $h$ durability and $s$ shield capacity. When Monocarp shoots an enemy spaceship, it receives $(P−s)$ damage (i. e. $(P−s)$ gets subtracted from its durability), where $P$ is the total power of the lasers that Monocarp shoots (i. e. $p_i$ if he only shoots laser $i$ and $p_1+p_2$ if he shoots both lasers at the same time). An enemy spaceship is considered destroyed when its durability becomes $0$ or lower.

Initially, both lasers are zero charged.

What's the lowest amount of time it can take Monocarp to destroy an enemy spaceship?

Input

The first line contains two integers $p_1$ and $t_1$ $(2 \leq p_1 \leq 5000; 1 \leq t_1 \leq {10}^{12})$ — the power and the reload time of the first laser.

The second line contains two integers $p_2$ and $t_2$ $(2 \leq p_2 \leq 5000; 1 \leq t_2 \leq {10}^{12})$ — the power and the reload time of the second laser.

The third line contains two integers $h$ and $s$ $(1 \leq h \leq 5000; 1 \leq s < \min(p_1,p_2))$ — the durability and the shield capacity of an enemy spaceship. Note that the last constraint implies that Monocarp will always be able to destroy an enemy spaceship.

Output

Print a single integer — the lowest amount of time it can take Monocarp to destroy an enemy spaceship.

Examples

input

5 10
4 9
16 1

output

20

input

10 1
5000 100000
25 9

output

25

 

解题思路

　　当时比赛时以为这题是贪心，后面看题解才知道是动态规划。看题解的时候也完全看不懂别人在写什么。今天再看了看大概明白了怎么一回事，写下来记录一下。

　　首先为什么可以用动态规划来做，是因为方案由激光射击的不同顺序组成，因此对于同一个的伤害，就有多种不同的激光射击顺序（对应的有不同的消耗时间），因此可以根据造成的伤害来定义状态，然后根据最后一次射击的激光来划分集合（单独射击激光$1$、单独射击激光$2$，两个激光同时射击）。

　　定义状态$f(i)$，表示所有造成至少$i$点伤害的激光射击方案，属性是消耗时间的最小值。如果最后一次是单个激光的射击，那么状态转移方程就比较好写。

　　如果最后一次射击是激光$1$，那么$f(i) = \min\{ {f(i), \,\, f(\max\{ {0, i - p_1 + s} \}) + t_1} \}$。

　　同理如果是最后一次射击是激光$2$，那么$f(i) = \min\{ {f(i), \,\, f(\max\{ {0, i - p_2 + s} \}) + t_2} \}$.

　　如果最后一次是同时射击，状态就不容易转移了，因为我们不知道加多少时间，即$f(i - p_1 - p_2 + s) + ?$。为了能求出这种状态的转移方程，我们需要明确的时间，假设同时射击前（包括）激光$1$一共射击了$j$次，那么在激光$1$射击了$j$次这段时间内（$j \times t_1$），激光$1$造成的伤害为$(j - 1) \times (p_1 - s)$，而激光$2$最多能造成的伤害为$\left\lfloor {(j \times t_1 - t_2) / t_2} \right\rfloor \times (p_2 - s)$，因此总的伤害就为$t = (j - 1) \times (p_1 - s) + \left\lfloor {(j \times t_1 - t_2) / t_2} \right\rfloor \times (p_2 - s) + p_1 + p_2 - s$，对应状态转方程就是$f(i) = \min\{ {f(i), \,\, f(\max\{ {0, i - t} \}) + j \times t_1} \}$。因此我们从$1$开始枚举$j$，由于一次攻击至少一点伤害，因此可以枚举到$i$。

　　同理也要枚举激光$2$射击的次数$j$，总伤害就是$t = (j - 1) \times (p_2 - s) + \left\lfloor {(j \times t_2 - t_1) / t_1} \right\rfloor \times (p_1 - s) + p_1 + p_2 - s$，状态转方程为$f(i) = \min\{ {f(i), \,\, f(\max\{ {0, i - t} \}) + j \times t_2} \}$。

　　AC代码如下，时间复杂度为$O(n^2)$：

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;

const int N = 5010;

LL f[N];

int main() {
    LL p1, t1, p2, t2, n, s;
    cin >> p1 >> t1 >> p2 >> t2 >> n >> s;
    
    memset(f, 0x3f, sizeof(f));
    f[0] = 0;
    for (int i = 1; i <= n; i++) {
        f[i] = min(f[max(0ll, i - p1 + s)] + t1, f[max(0ll, i - p2 + s)] + t2);    // 最后一次是单独射击的情况
        // 最后一次是同时射击的情况
        for (int j = 1; j <= i; j++) {
            if (j * t1 < t2) continue;    // 因为同时射击，因此激光2至少要射击一次
            LL t = (j - 1) * (p1 - s) + (j * t1 - t2) / t2 * (p2 - s) + p1 + p2 - s;
            f[i] = min(f[i], f[max(0ll, i - t)] + j * t1);
        }
        for (int j = 1; j <= i; j++) {
            if (j * t2 < t1) continue;    // 因为同时射击，因此激光1至少要射击一次
            LL t = (j - 1) * (p2 - s) + (j * t2 - t1) / t1 * (p1 - s) + p1 + p2 - s;
            f[i] = min(f[i], f[max(0ll, i - t)] + j * t2);
        }
    }
    
    cout << f[n];
    
    return 0;
}

 

参考资料

　　Educational Codeforces Round 137 (Rated for Div. 2) 补D、E：http://www.wtld.cn/a/72097.html