Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals) D

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Joky_2002

最新推荐文章于 2022-08-21 01:05:54 发布

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分类专栏： Codeforces 贪心

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解决一组人员如何在有限数量的钥匙中找到最佳路径方案以最短时间抵达办公室的问题，涉及二分查找及贪心算法。

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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 per 1 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 and p (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 integers a1, 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 integers b1, 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 point 40 and go with it to the office located at point 50. He spends 30 seconds. The person located at point 100 can take the key located at point 80 and go to the office with it. He spends 50seconds. Thus, after 50 seconds everybody is in office with keys.

好傻逼的题目。。。。。

考虑二分+贪心，根据排序不等式，尽量在当前答案下往左取才是最优的

代码：

#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<vector>
using namespace std;

typedef double DL;

typedef long long LL;

const int maxn = 3000;
const int INF = 2147483647;

int a[maxn],b[maxn],n,k,p,ans;

inline int getint()
{
	int ret = 0,f = 1;
	char c = getchar();
	while (c < '0' || '9' < c) 
	{
		if (c == '-') f = -1;
		c = getchar();
	}
	while ('0' <= c && c <= '9')
		ret = ret * 10 + c - '0',c = getchar();
	return ret * f;
}

inline bool judge(int x)
{
	int j = 1;
	for (int i = 1; i <= n; i++)
	{
		while (j <= k && abs(a[i] - b[j]) + abs(b[j] - p) > x) j++;
		if (j > k) return 0;
		j++;
	}
	return 1;
}

int main()
{
	#ifdef AMC
		freopen("AMC1.txt","r",stdin);
	#endif
	n = getint(); k = getint(); p = getint();
	for (int i = 1; i <= n; i++) a[i] = getint();
	sort(a + 1,a + n + 1);
	for (int i = 1; i <= k; i++) b[i] = getint();
	sort(b + 1,b + k + 1);
	LL l = 0,r = INF;
	while (r - l > 1)
	{
		int mid = l + r >> 1;
		if (judge(mid)) r = mid;
		else l = mid;
	}
	if (judge(l)) ans = l;
	else ans = r;
	printf("%d",ans);
	return 0;
}