题目链接

https://arc073.contest.atcoder.jp/tasks/arc073_d

题意简述

有nnn个格子，有两颗棋子，初始位置为A,BA,BA,B，移动一颗棋子从XXX到YYY的代价是∣X−Y∣|X-Y|∣X−Y∣，有QQQ个要求，每个要求是将任意一颗棋子移动到XiX_iXi​点，求代价最小值。

题解

设f[i][j]f[i][j]f[i][j]表示前iii个要求被满足，一颗棋子在XiX_iXi​，另一颗在jjj花费的最小代价，转移
f[i][j]=f[i−1][j]+∣Xi−Xi−1∣f[i][Xi−1]=min⁡k=1n(f[i−1][k]+∣k−Xi∣) f[i][j]=f[i-1][j]+|X_i-X_{i-1}|\\ f[i][X_{i-1}]=\min_{k=1}^{n} (f[i-1][k]+|k-X_i|) f[i][j]=f[i−1][j]+∣Xi​−Xi−1​∣f[i][Xi−1​]=k=1minn​(f[i−1][k]+∣k−Xi​∣)
发现第一维可以省掉，用线段树维护第二维，维护三个值(f[i]+i),(f[i]+n−i+1),f[i](f[i]+i),(f[i]+n-i+1),f[i](f[i]+i),(f[i]+n−i+1),f[i]，第一种转移就是区间加，第二种就只要对Xi−1X_{i-1}Xi−1​右侧的(f[i]+i)−Xi(f[i]+i)-X_i(f[i]+i)−Xi​，左侧的(f[i]+n−i+1)−n+Xi−1(f[i]+n-i+1)-n+X_i-1(f[i]+n−i+1)−n+Xi​−1取min⁡\minmin，然后区间修改就行了。

代码

#include <cstdio>
#include <algorithm>int read()
{int x=0,f=1;char ch=getchar();while((ch<'0')||(ch>'9')){if(ch=='-'){f=-f;}ch=getchar();}while((ch>='0')&&(ch<='9')){x=x*10+ch-'0';ch=getchar();}return x*f;
}const int maxn=200000;
const long long inf=0x3f3f3f3f3f3f3f3fll;int n,q,sta,stb,a[maxn+10];namespace sgt
{long long mn[3][maxn<<2],adt[maxn<<2];int puttag(int x,long long v){mn[0][x]+=v;mn[1][x]+=v;mn[2][x]+=v;adt[x]+=v;return 0;}int pushdown(int x){puttag(x<<1,adt[x]);puttag(x<<1|1,adt[x]);adt[x]=0;return 0;}int updata(int x){mn[0][x]=std::min(mn[0][x<<1],mn[0][x<<1|1]);mn[1][x]=std::min(mn[1][x<<1],mn[1][x<<1|1]);mn[2][x]=std::min(mn[2][x<<1],mn[2][x<<1|1]);return 0;}int build(int x,int l,int r){if(l==r){mn[0][x]=inf+l;mn[1][x]=inf+n-l+1;mn[2][x]=inf;return 0;}int mid=(l+r)>>1;build(x<<1,l,mid);build(x<<1|1,mid+1,r);updata(x);return 0;}int add(int x,int l,int r,int askl,int askr,long long adv){if((askl<=l)&&(r<=askr)){puttag(x,adv);return 0;}int mid=(l+r)>>1;pushdown(x);if(askl<=mid){add(x<<1,l,mid,askl,askr,adv);}if(mid<askr){add(x<<1|1,mid+1,r,askl,askr,adv);}updata(x);return 0;}long long getmn(int x,int l,int r,int askl,int askr,int id){if((askl<=l)&&(r<=askr)){return mn[id][x];}int mid=(l+r)>>1;long long res=inf<<1;pushdown(x);if(askl<=mid){res=std::min(res,getmn(x<<1,l,mid,askl,askr,id));}if(mid<askr){res=std::min(res,getmn(x<<1|1,mid+1,r,askl,askr,id));}return res;}
}int abs(int x)
{return (x<0)?(-x):x;
}int main()
{n=read();q=read();sta=read();stb=read();for(int i=1; i<=q; ++i){a[i]=read();}a[0]=sta;sgt::build(1,1,n);sgt::add(1,1,n,stb,stb,-inf);for(int i=1; i<=q; ++i){long long oth=std::min(sgt::getmn(1,1,n,1,a[i],1)-(n-a[i]+1),sgt::getmn(1,1,n,a[i],n,0)-a[i]);sgt::add(1,1,n,1,n,abs(a[i]-a[i-1]));long long nth=sgt::getmn(1,1,n,a[i-1],a[i-1],2);if(nth>oth){sgt::add(1,1,n,a[i-1],a[i-1],oth-nth);}}printf("%lld\n",sgt::getmn(1,1,n,1,n,2));return 0;
}