CF709B Checkpoints 模拟

最新推荐文章于 2024-08-14 17:01:39 发布

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最新推荐文章于 2024-08-14 17:01:39 发布

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原文链接：http://www.cnblogs.com/zxyqzy/p/10127178.html

本文探讨了在一次定向越野比赛中，参赛者Vasya如何从起点出发，以最小化总行走距离的方式访问至少n-1个检查点。通过分析不同情况，文章提供了一种有效的算法来计算最小可能的行走距离。

Vasya takes part in the orienteering competition. There are n checkpoints located along the line at coordinates x₁, x₂, ..., x_n. Vasya starts at the point with coordinate a. His goal is to visit at least n - 1 checkpoint in order to finish the competition. Participant are allowed to visit checkpoints in arbitrary order.

Vasya wants to pick such checkpoints and the order of visiting them that the total distance travelled is minimized. He asks you to calculate this minimum possible value.

Input

The first line of the input contains two integers n and a (1 ≤ n ≤ 100 000, - 1 000 000 ≤ a ≤ 1 000 000) — the number of checkpoints and Vasya's starting position respectively.

The second line contains n integers x₁, x₂, ..., x_n ( - 1 000 000 ≤ x_i ≤ 1 000 000) — coordinates of the checkpoints.

Output

Print one integer — the minimum distance Vasya has to travel in order to visit at least n - 1 checkpoint.

Examples

Input

Copy

3 10
1 7 12

Output

Copy

Input

Copy

2 0
11 -10

Output

Copy

Input

Copy

5 0
0 0 1000 0 0

Output

Copy

Note

In the first sample Vasya has to visit at least two checkpoints. The optimal way to achieve this is the walk to the third checkpoints (distance is 12 - 10 = 2) and then proceed to the second one (distance is 12 - 7 = 5). The total distance is equal to 2 + 5 = 7.

In the second sample it's enough to visit only one checkpoint so Vasya should just walk to the point - 10.

4种情况求一下min即可；

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 1000005
#define inf 0x3f3f3f3f
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
    ll x = 0;
    char c = getchar();
    bool f = false;
    while (!isdigit(c)) {
        if (c == '-') f = true;
        c = getchar();
    }
    while (isdigit(c)) {
        x = (x << 1) + (x << 3) + (c ^ 48);
        c = getchar();
    }
    return f ? -x : x;
}

ll gcd(ll a, ll b) {
    return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }

/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
    if (!b) {
        x = 1; y = 0; return a;
    }
    ans = exgcd(b, a%b, x, y);
    ll t = x; x = y; y = t - a / b * y;
    return ans;
}
*/

int n, a;
ll x[maxn];


int main()
{
    //ios::sync_with_stdio(0);
    rdint(n); rdint(a);
    for (int i = 1; i <= n; i++)rdllt(x[i]);
    ll sum = 0;
    sort(x + 1, x + 1 + n);
    if (n == 1)cout << 0 << endl;
    else {
        sum = inf;
        sum = min(sum, abs(a - x[1]) + x[n - 1] - x[1]);
        sum = min(sum, abs(a - x[n - 1]) + x[n - 1] - x[1]);
        sum = min(sum, abs(a - x[n]) + x[n] - x[2]);
        sum = min(sum, abs(a - x[2]) + x[n] - x[2]);
        cout << sum << endl;
    }
    return 0;
}

转载于:https://www.cnblogs.com/zxyqzy/p/10127178.html