F

F - Many Moves

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Time limit : 2sec / Memory limit : 256MB

Score : 900 points

Problem Statement

There are N squares in a row. The squares are numbered 1,2,…,N from left to right.

You have two pieces, initially placed on square A and B, respectively. You will be asked to process Q queries of the following kind, in the order received:

• Given an integer xi, move one of the two pieces of your choice to square xi.

Here, it takes you one second to move a piece one square. That is, the time it takes to move a piece from square X to Y is |X−Y| seconds.

Your objective is to process all the queries in the shortest possible time.

You may only move the pieces in response to queries, and you may not move both pieces at the same time. Also, it is not allowed to rearrange the order in which queries are given. It is, however, allowed to have both pieces in the same square at the same time.

Constraints

• 1≤N,Q≤200,000
• 1≤A,B≤N
• 1≤xi≤N

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Input

Input is given from Standard Input in the following format:

N Q A B
x1 x2 ... xQ

Output

Let the shortest possible time to process all the queries be X seconds. Print X.

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Sample Input 1

8 3 1 8
3 5 1

Sample Output 1

7

All the queries can be processed in seven seconds, by:

• moving the piece at square 1 to 3
• moving the piece at square 8 to 5
• moving the piece at square 3 to 1

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Sample Input 2

9 2 1 9
5 1

Sample Output 2

4

The piece at square 9 should be moved first.

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Sample Input 3

9 2 1 9
5 9

Sample Output 3

4

The piece at square 1 should be moved first.

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Sample Input 4

11 16 8 1
1 1 5 1 11 4 5 2 5 3 3 3 5 5 6 7

Sample Output 4

21
分析：考虑dp[i][j]表示当前在x[i],j位置；
　　　设之前一步在a，b，当前到c，d，且a，c为上次和这次到达点；
　　　那么有a->c或b->c；
　　　若a->c，则dp[i][j]直接加上abs(x[i]-x[i-1]);
　　　若b->c，则dp[i][a]取min{dp[i-1][j]+abs(j-x[i])};
　　　而这两个都可以用线段树维护；
代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <bitset>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <cassert>
#include <ctime>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define mod 1000000009
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define sys system("pause")
#define ls rt<<1
#define rs rt<<1|1
const int maxn=2e5+10;
const int N=2e5+10;
using namespace std;
int id(int l,int r){return l+r|l!=r;}
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p%mod;p=p*p%mod;q>>=1;}return f;}
int n,m,k,t,q,a,b;
ll tag[maxn<<2],mi[maxn<<2],mi1[maxn<<2],mi2[maxn<<2];
void pup(int rt)
{
    mi[rt]=min(mi[ls],mi[rs]);
    mi1[rt]=min(mi1[ls],mi1[rs]);
    mi2[rt]=min(mi2[ls],mi2[rs]);
    tag[rt]=0;
}
void pdw(int rt)
{
    mi[ls]+=tag[rt];
    mi1[ls]+=tag[rt];
    mi2[ls]+=tag[rt];
    tag[ls]+=tag[rt];
    mi[rs]+=tag[rt];
    mi1[rs]+=tag[rt];
    mi2[rs]+=tag[rt];
    tag[rs]+=tag[rt];
    tag[rt]=0;
}
void build(int l,int r,int rt)
{
    if(l==r)
    {
        mi[rt]=mi1[rt]=mi2[rt]=1e18;
        tag[rt]=0;
        return;
    }
    int mid=l+r>>1;
    build(l,mid,ls);
    build(mid+1,r,rs);
    pup(rt);
}
void add(int L,int R,ll v,int l,int r,int rt)
{
    if(L==l&&R==r)
    {
        mi[rt]+=v;
        mi1[rt]+=v;
        mi2[rt]+=v;
        tag[rt]+=v;
        return;
    }
    int mid=l+r>>1;
    if(tag[rt])pdw(rt);
    if(R<=mid)add(L,R,v,l,mid,ls);
    else if(L>mid)add(L,R,v,mid+1,r,rs);
    else
    {
        add(L,mid,v,l,mid,ls);
        add(mid+1,R,v,mid+1,r,rs);
    }
    pup(rt);
}
void upd(int pos,ll v,int l,int r,int rt)
{
    if(l==pos&&pos==r)
    {
        if(mi[rt]>v)
        {
            mi[rt]=v;
            mi1[rt]=v-pos;
            mi2[rt]=v+pos;
        }
        return;
    }
    int mid=l+r>>1;
    if(tag[rt])pdw(rt);
    if(pos<=mid)upd(pos,v,l,mid,ls);
    else upd(pos,v,mid+1,r,rs);
    pup(rt);
}
ll gao(int L,int R,int l,int r,int rt,ll *mi)
{
    if(L==l&&R==r)return mi[rt];
    int mid=l+r>>1;
    if(tag[rt])pdw(rt);
    if(R<=mid)return gao(L,R,l,mid,ls,mi);
    else if(L>mid)return gao(L,R,mid+1,r,rs,mi);
    else return min(gao(L,mid,l,mid,ls,mi),gao(mid+1,R,mid+1,r,rs,mi));
}
int main()
{
    int i,j;
    scanf("%d%d%d%d",&n,&q,&a,&b);
    build(1,n,1);
    upd(b,0,1,n,1);
    int pre=a;
    rep(i,1,q)
    {
        int x;
        scanf("%d",&x);
        ll cost1=gao(1,x,1,n,1,mi1)+x;
        ll cost2=gao(x,n,1,n,1,mi2)-x;
        ll now=min(cost1,cost2);
        add(1,n,abs(x-pre),1,n,1);
        upd(pre,now,1,n,1);
        pre=x;
    }
    printf("%lld
",gao(1,n,1,n,1,mi));
    return 0;
}