排序机械臂

题目大意

给出一个长度为n的序列a，有n次操作，第i次操作把数值a第i小的位置记作p[i]，位置i到p[i]的数反转（如1,2,3,4反转后变成4,3,2,1）。求输出序列p。

范围a[i]<=2*10^9,n<=10^5。

先离散化

裸splay

用splay树维护size。答案p[i]=size[i的左儿子]+1。
对于反转操作，可以打个标记。

代码

#include<cstdio>#include<cstring>#include<algorithm>#define fo(i,a,b) for(i=a;i<=b;i++)using namespace std;const int maxn=100000+5;struct ar{    int x,y;} a[maxn];int fa[maxn],b[maxn],size[maxn],tree[maxn][2],re[maxn],q[maxn];int n,i,j;bool cmp(ar a,ar b) {    return a.x<b.x||((a.x==b.x)&&(a.y<b.y));}int pd(int x){    if (tree[fa[x]][0]==x) return 0;return 1;}void update(int x) {    size[x]=size[tree[x][0]]+size[tree[x][1]]+1;}void clear(int x){    int l=tree[x][0],r=tree[x][1];    if (re[x]==0) return;    re[x]=0;    re[l]=1-re[l],re[r]=1-re[r];    swap(tree[x][0],tree[x][1]);}void chu(int x,int y){    q[0]=0;    while (x!=y){        q[++q[0]]=x;        x=fa[x];    }    int i;    fo(i,1,q[0]) clear(q[q[0]-i+1]);}void rotate(int x){    int y=fa[x],z=pd(x);    fa[x]=fa[y];    if (fa[y]) tree[fa[y]][pd(y)]=x;    tree[y][z]=tree[x][1-z];    if (tree[x][1-z]) fa[tree[x][1-z]]=y;    tree[x][1-z]=y;    fa[y]=x;    update(y),update(x);}void splay(int x,int y){    chu(x,y);    while (fa[x]!=y){        int f=fa[x];        if (fa[f]!=y)         if (pd(x)==pd(f)) rotate(f);else rotate(x);        rotate(x);    }}int kth(int x,int y){    clear(x);    if (size[tree[x][0]]+1==y) return x;    if (size[tree[x][0]]+1>y) return kth(tree[x][0],y);    return kth(tree[x][1],y-size[tree[x][0]]-1);}void split(int x,int y,int &l,int &r){    int j=kth(x,y);    splay(j,0);    l=j,r=tree[j][1];    tree[l][1]=0;    fa[r]=0;    update(j);}void merge(int x,int y,int &l){    int j=kth(x,size[x]);    splay(j,0);    tree[j][1]=y;    fa[y]=j;    update(j);    l=j;}int main(){    scanf("%d",&n);    fo(i,1,n) {        scanf("%d",&a[i].x);        a[i].y=i;    }    sort(a+1,a+n+1,cmp);    fo(i,1,n) b[a[i].y]=i;size[n+1]=size[n+2]=1;    tree[b[1]][0]=n+1,fa[n+1]=b[1],update(b[1]);    fo(i,2,n) {        tree[b[i]][0]=b[i-1];        fa[b[i-1]]=b[i];        update(b[i]);    }    tree[n+2][0]=b[n],fa[b[n]]=n+2,update(n+2);    fo(i,1,n-1){        int l,r,mid;        splay(i,0);        printf("%d ",size[tree[i][0]]);        if (i==n-1) continue;        split(i,size[tree[i][0]]+1,l,r);        split(l,i,l,mid);        re[mid]=1-re[mid];        merge(l,mid,l);        merge(l,r,l);    }    printf("%d",n);}