Codeforces 831D Office Keys【二分】

最新推荐文章于 2020-01-22 16:02:41 发布

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最新推荐文章于 2020-01-22 16:02:41 发布

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解决N个人使用K把钥匙抵达终点的问题，通过排序与二分查找确定所有人拿到钥匙并到达终点所需的最短时间。

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There are n people and k keys on a straight line. Every person wants to get to the office which is located on the line as well. To do that, he needs to reach some point with a key, take the key and then Go to the office. Once a key is taken by somebody, it couldn't be taken by anybody else.

You are to determine the minimum time needed for all n people to get to the office with keys. Assume that people move a unit distance per1 second. If two people reach a key at the same time, only one of them can take the key. A person can pass through a point with a key without taking it.

Input

The first line contains three integers n,k andp (1 ≤ n ≤ 1 000,n ≤ k ≤ 2 000,1 ≤ p ≤ 10⁹) — the number of people, the number of keys and the office location.

The second line contains n distinct integersa₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹) — positions in which people are located initially. The positions are given in arbitrary order.

The third line contains k distinct integersb₁, b₂, ..., b_k (1 ≤ b_j ≤ 10⁹) — positions of the keys. The positions are given in arbitrary order.

Note that there can't be more than one person or more than one key in the same point. A person and a key can be located in the same point.

Output

Print the minimum time (in seconds) needed for all n to reach the office with keys.

Examples

Input

2 4 50
20 100
60 10 40 80

Output

Input

1 2 10
11
15 7

Output

Note

In the first example the person located at point 20 should take the key located at point40 and go with it to the office located at point50. He spends 30seconds. The person located at point100 can take the key located at point80 and go to the office with it. He spends 50 seconds. Thus, after50 seconds everybody is in office with keys.

题目大意：

给你N个人，有K把钥匙，终点为P.

每个人要求拿一把钥匙到终点，问多长时间，能够使得所有人都做到。

思路：先按照大小排序，然后二分答案，判断是否可行

#include <iostream>
#include <algorithm>
#include <string.h>
#include <stdio.h>
#include <cmath>
#include <vector>
#include <utility>
#include <queue>
using namespace std;
int a[1005];
int b[2005];
int visited[2005];
int n, k, p;
bool bi(long long num) {
    memset(visited, 0, sizeof(visited));
    int cnt = 0;
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= k; ++j) {
            if (visited[j])
                continue;
            if (abs(a[i] - b[j]) + abs(b[j] - p) <= num) {
                cnt++;
                visited[j] = 1;
                break;
            }
        }
    }
    return cnt == n;
}
int main() {
        //freopen("in.txt", "r", stdin);
    scanf("%d%d%d", &n, &k, &p);
    
    for (int i = 1; i <= n; ++i) {
        scanf("%d", &a[i]);
    }
    for (int i = 1; i <= k; ++i) {
        scanf("%d", &b[i]);
    }
    sort(a + 1, a + n + 1, greater<int>());
    sort(b + 1, b + k + 1, greater<int>());
    long long left = 0, right = 2e9;
    while (left < right) {
        long long mid = (left + right) >> 1;
            //cout << mid << endl;
        if (bi(mid)) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    cout << right << endl;
}