题意：在一维坐标系下，给定n个草坪的坐标(1<=n<=1000)，再给定起点L(1<=L<=1e6)。每个草坪有一个staleness值，初始化为0。每次走一格，所有的staleness值+1，并且要从起点遍历每一个草坪，求staleness值的最小值。

离散化处理n个点
dp[i][j][0]:区间[i, j]已经遍历完，目前在i点时，staleness的最小值。
dp[i][j][1]:区间[i, j]已经遍历完，目前在j点时，staleness的最小值。
在更新dp[i][j][0/1]的过程中，有n-(j-i)个点的staleness在变化
dp[i][j][0]:
(1)从[i+1, j]的左端点到i：dp[i+1][j][0]+(n-(j-i))*(a[i+1]-a[i]);
(2)从[i+1, j]的右端点到i：dp[i+1][j][1]+(n-(j-i))*(a[j]-a[i]);
dp[i][j][1]：
(1)从[i, j-1]的左端点到j：dp[i][j-1][0]+(n-(j-i))*(a[j]-a[i]));
(2)从[i, j-1]的右端点到j：dp[i][j-1][1]+(n-(j-i))*(a[j]-a[j-1]));

#include <cstdio>
#include <algorithm>
#include <cstring>

using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 1010;
int a[N];
int dp[N][N][2];
int n, L;

int bin(int l, int r, int key)
{
    while(l<=r)
    {
        int mid = (l+r)/2;
        if(a[mid]>=key) r = mid-1;
        else l = mid+1;
    }
    return r+1;
}

int main()
{
    scanf("%d %d", &n, &L);
    bool flag = 1;
    for(int i = 1; i <= n; ++i)
    {
        scanf("%d", &a[i]);
        if(a[i]==L) flag = 0;
    }
    if(flag) a[++n] = L;
    sort(a+1, a+n+1);
    int pos = bin(1, n, L);
    memset(dp, INF, sizeof(dp));
    dp[pos][pos][0] = dp[pos][pos][1] = 0;
    for(int i = pos; i >= 1; --i)
        for(int j = pos; j <= n; ++j)
        {
            dp[i][j][0] = min(dp[i][j][0], dp[i+1][j][0]+(n-(j-i))*(a[i+1]-a[i]));
            dp[i][j][0] = min(dp[i][j][0], dp[i+1][j][1]+(n-(j-i))*(a[j]-a[i]));
            dp[i][j][1] = min(dp[i][j][1], dp[i][j-1][0]+(n-(j-i))*(a[j]-a[i]));
            dp[i][j][1] = min(dp[i][j][1], dp[i][j-1][1]+(n-(j-i))*(a[j]-a[j-1]));
        }
    printf("%d\n", min(dp[1][n][0], dp[1][n][1]));
    return 0;
}