Codeforces Round #379 (Div. 2) C. Anton and Making Potions(二分）

Posted 2020-08-19 Gealo

Anton is playing a very interesting computer game, but now he is stuck at one of the levels. To pass to the next level he has to prepare npotions.

Anton has a special kettle, that can prepare one potions in x seconds. Also, he knows spells of two types that can faster the process of preparing potions.

1. Spells of this type speed up the preparation time of one potion. There are m spells of this type, the i-th of them costs bi manapoints and changes the preparation time of each potion to ai instead of x. 
2. Spells of this type immediately prepare some number of potions. There are k such spells, the i-th of them costs di manapoints and instantly create ci potions. 

Anton can use no more than one spell of the first type and no more than one spell of the second type, and the total number of manapoints spent should not exceed s. Consider that all spells are used instantly and right before Anton starts to prepare potions.

Anton wants to get to the next level as fast as possible, so he is interested in the minimum number of time he needs to spent in order to prepare at least n potions.

Input

The first line of the input contains three integers n, m, k (1 ≤ n ≤ 2·109, 1 ≤ m, k ≤ 2·105) — the number of potions, Anton has to make, the number of spells of the first type and the number of spells of the second type.

The second line of the input contains two integers x and s (2 ≤ x ≤ 2·109, 1 ≤ s ≤ 2·109) — the initial number of seconds required to prepare one potion and the number of manapoints Anton can use.

The third line contains m integers ai (1 ≤ ai < x) — the number of seconds it will take to prepare one potion if the i-th spell of the first type is used.

The fourth line contains m integers bi (1 ≤ bi ≤ 2·109) — the number of manapoints to use the i-th spell of the first type.

There are k integers ci (1 ≤ ci ≤ n) in the fifth line — the number of potions that will be immediately created if the i-th spell of the second type is used. It‘s guaranteed that ci are not decreasing, i.e. ci ≤ cj if i < j.

The sixth line contains k integers di (1 ≤ di ≤ 2·109) — the number of manapoints required to use the i-th spell of the second type. It‘s guaranteed that di are not decreasing, i.e. di ≤ dj if i < j.

Output

Print one integer — the minimum time one has to spent in order to prepare n potions.

 

题意：要求得到至少n个药剂，可以使用两种魔法，一种能够缩短制药时间，一种能瞬间制药，

给你x表示标准制药一个要x秒，给你s表示你的法力值为s

m种第一类类魔法，消耗b点魔法，缩短时间为a秒。

k种第二类魔法，消耗d点魔法，瞬间做出c个药。

两种魔法最多各选一个用，问你最少花多少时间能制得至少n个药剂

由于题目给出的c，d是递增的，所以这题相对比较简单，只要遍历一遍第一类魔法再二分查找一下最大且和不超过s的第二类魔法这样就能确保

找到的是最优解，有点贪心的思想。还有一点要注意的，最优的选择可以不用魔法，或者只用一种魔法，这个要注意一下的。

 

#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <cstdio>
using namespace std;
typedef long long ll;
const int M = 2e5 + 20;
ll a[M] , b[M] , c[M] , d[M];
int main()
{
    ll n , m , k;
    scanf("%I64d%I64d%I64d" , &n , &m , &k);
    ll x , s;
    scanf("%I64d%I64d" , &x , &s);
    for(int i = 0 ; i < m ; i++) {
        scanf("%I64d" , &a[i]);
    }
    for(int i = 0 ; i < m ; i++) {
        scanf("%I64d" , &b[i]);
    }
    for(int i = 0 ; i < k ; i++) {
        scanf("%I64d" , &c[i]);
    }
    for(int i = 0 ; i < k ; i++) {
        scanf("%I64d" , &d[i]);
    }
    ll MIN = n * x;
    a[m] = x;
    for(int i = 0 ; i <= m ; i++) {
        if(s >= b[i]) {
            ll temp = s - b[i];
            int pos = upper_bound(d , d + k , temp) - d;
            if(pos == 0) {
                MIN = min(MIN , n * a[i]);
                continue;
            }
            pos--;
            ll gg = n - c[pos];
            gg *= a[i];
            MIN = min(MIN , gg);
        }
    }
    printf("%I64d\n" , MIN);
    return 0;
}