Codeforces 831D Office Keys[2점]

D. Office Keys
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
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, ..., an (1 ≤ ai ≤ 109) — positions in which people are located initially. The positions are given in arbitrary order.
The third line contains k distinct integersb1, b2, ..., bk (1 ≤ bj ≤ 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

50

Input

1 2 10
11
15 7

Output

7

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.
제목 대의:
N 명 드릴게요. K 열쇠가 있어요. 종점은 P예요.
모든 사람이 열쇠를 가지고 결승점까지 얼마나 걸릴지 물어보면 모든 사람이 할 수 있다.
사고방식: 먼저 크기에 따라 순서를 정한 다음에 답을 나누어 실행 여부를 판단한다.

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
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());
    sort(b + 1, b + k + 1, greater());
    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;
}