poj3580 splay树 REVOVLE循环

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SuperMemo

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 12795 Accepted: 3989Case Time Limit: 2000MS

Description

Your friend, Jackson is invited to a TV show called SuperMemo in which the participant is told to play a memorizing game. At first, the host tells the participant a sequence of numbers, {A₁, A₂, ... A_n}. Then the host performs a series of operations and queries on the sequence which consists:

ADD x y D: Add D to each number in sub-sequence {A_x ... A_y}. For example, performing "ADD 2 4 1" on {1, 2, 3, 4, 5} results in {1, 3, 4, 5, 5}
REVERSE x y: reverse the sub-sequence {A_x ... A_y}. For example, performing "REVERSE 2 4" on {1, 2, 3, 4, 5} results in {1, 4, 3, 2, 5}
REVOLVE x y T: rotate sub-sequence {A_x ... A_y} T times. For example, performing "REVOLVE 2 4 2" on {1, 2, 3, 4, 5} results in {1, 3, 4, 2, 5}
INSERT x P: insert P after A_x. For example, performing "INSERT 2 4" on {1, 2, 3, 4, 5} results in {1, 2, 4, 3, 4, 5}
DELETE x: delete A_x. For example, performing "DELETE 2" on {1, 2, 3, 4, 5} results in {1, 3, 4, 5}
MIN x y: query the participant what is the minimum number in sub-sequence {A_x ... A_y}. For example, the correct answer to "MIN 2 4" on {1, 2, 3, 4, 5} is 2

To make the show more interesting, the participant is granted a chance to turn to someone else that means when Jackson feels difficult in answering a query he may call you for help. You task is to watch the TV show and write a program giving the correct answer to each query in order to assist Jackson whenever he calls.

Input

The first line contains n (n ≤ 100000).

The following n lines describe the sequence.

Then follows M (M ≤ 100000), the numbers of operations and queries.

The following M lines describe the operations and queries.

Output

For each "MIN" query, output the correct answer.

Sample Input

51 2 3 4 52ADD 2 4 1MIN 4 5

Sample Output

/*poj3580 splay树 REVOVLE循环给定一个数列:a1,a2,.... an进行以下6种操作：ADD x y D : 给第x个数到第y个数加DREVERSE x y : 反转[x,y]REVOVLE x y T : 对[x,y]区间的数循环右移T次  (这个最开始没想到这么弄)(先把T对长度取模，然后相当于把[y-T+1,y]放到[x,y-T] 的前面，T可能出现负数或者特别大)INSERT x P  : 在第x个数后面插入PDELETE x : 删除第x个数 MIN x y  : 查询[x,y]之间的最小的数像min，add是参照rev，size来写。然后就只有第三个操作可以一开始想不到，其它大致就是各种基本操作的组合了最开始写错del导致TLE几次hhh-2016-02-21 03:06:02*/#include <functional>#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <map>#include <cmath>using namespace std;typedef long long ll;typedef long  double ld;#define key_value ch[ch[root][1]][0]const int maxn = 200010;int ch[maxn][2];int pre[maxn],siz[maxn],num[maxn];int rev[maxn],key[maxn];int add[maxn];int Min[maxn],a[maxn];int tot;int root;void push_up(int r){    int lson = ch[r][0],rson = ch[r][1];    siz[r] = siz[lson] + siz[rson] + 1;    Min[r] = min(key[r],min(Min[lson],Min[rson]));}void update_add(int r,int val){    if(!r) return;    key[r] += val;    add[r] += val;    Min[r] += val;}void update_rev(int r){    if(!r)return ;    swap(ch[r][0],ch[r][1]);    rev[r] ^= 1;}void push_down(int r){    if(rev[r])    {        update_rev(ch[r][0]);        update_rev(ch[r][1]);        rev[r] = 0;    }    if(add[r])    {        update_add(ch[r][0],add[r]);        update_add(ch[r][1],add[r]);        add[r] = 0;    }}void NewNode(int &r,int far,int k){    r = ++tot;      //不能为0    pre[r] = far;    ch[r][0] = ch[r][1] = 0;    siz[r] = 1;    Min[r] = k;    key[r] = k;    rev[r] = 0;    add[r] = 0;}void rotat(int x,int kind){    int y = pre[x];    push_down(y);    push_down(x);    ch[y][!kind] = ch[x][kind];    pre[ch[x][kind]] = y;    if(pre[y])        ch[pre[y]][ch[pre[y]][1]==y] = x;    pre[x] = pre[y];    ch[x][kind] = y;    pre[y] = x;    push_up(y);}void build(int &x,int l,int r,int far){    if(l > r) return ;    int mid = (l+r) >>1;    NewNode(x,far,a[mid]);    build(ch[x][0],l,mid-1,x);    build(ch[x][1],mid+1,r,x);    push_up(x);}void splay(int r,int goal){    push_down(r);    while(pre[r] != goal)    {        if(pre[pre[r]] == goal)        {            push_down(pre[r]);            push_down(r);            rotat(r,ch[pre[r]][0] == r);        }        else        {            push_down(pre[pre[r]]);            push_down(pre[r]);            push_down(r);            int y = pre[r];            int kind = ch[pre[y]][0] == y;            if(ch[y][kind] == r)            {                rotat(r,!kind);                rotat(r,kind);            }            else            {                rotat(y,kind);                rotat(r,kind);            }        }    }    push_up(r);    if(goal == 0)        root = r;}int get_kth(int r,int k){    push_down(r);    int t = siz[ch[r][0]] + 1;    if(k == t)return r;    if(t > k) return get_kth(ch[r][0],k);    else return get_kth(ch[r][1],k-t);}int get_next(int r){    push_down(r);    if(ch[r][1] == 0)return -1;    r = ch[r][1];    while(ch[r][0])    {        r = ch[r][0];        push_down(r);    }    return r;}void Reverse(int l,int r){    splay(get_kth(root,l),0);    splay(get_kth(root,r+2),root);    update_rev(key_value);    push_up(ch[root][1]);    push_up(root);}void Add(int l,int r,int val){    splay(get_kth(root,l),0);    splay(get_kth(root,r+2),root);    update_add(key_value,val);    push_up(ch[root][1]);    push_up(root);}void ini(int n){    tot = root = 0;    ch[root][0] = ch[root][1] = pre[root] = siz[root] = num[root] = 0;    Min[root] = 0x3f3f3f3f;    rev[root] = add[root] = 0;    NewNode(root,0,-1);    NewNode(ch[root][1],root,-1);    for(int i=1; i <= n; i++)    {        scanf("%d",&a[i]);    }    build(key_value,1,n,ch[root][1]);    push_up(ch[root][1]);    push_up(root);}int get_min(int r){    push_down(r);    while(ch[r][0])    {        r = ch[r][0];        push_down(r);    }    return r;}void Delete(int r){    splay(get_kth(root,r+1),0);    if(ch[root][0] == 0 || ch[root][1] == 0)    {        root = ch[root][0] + ch[root][1];        pre[root] = 0;        return;    }    int k = get_min(ch[root][1]);    splay(k,root);    ch[ch[root][1]][0] = ch[root][0];    root = ch[root][1];    pre[ch[root][0]] = root;    pre[root] = 0;    push_up(root);}void Insert(int x,int y){    splay(get_kth(root,x+1),0);    splay(get_kth(root,x+2),root);    NewNode(key_value,ch[root][1],y);    push_up(ch[root][1]);    push_up(root);}int MIN(int x,int y){    splay(get_kth(root,x),0);    splay(get_kth(root,y+2),root);    push_up(ch[root][1]);    push_up(root);    return Min[key_value];}void Revovle(int x,int y,int T){    splay(get_kth(root,y-T+1),0);    splay(get_kth(root,y+2),root);    int tmp = key_value;    //pre[key_value] = 0;    key_value = 0;    push_up(ch[root][1]);    push_up(root);    splay(get_kth(root,x),0);    splay(get_kth(root,x+1),root);    key_value = tmp;    pre[key_value] = ch[root][1];    push_up(ch[root][1]);    push_up(root);}int main(){    int n,p;    while(scanf("%d",&n) != EOF)    {        ini(n);        scanf("%d",&p);        char opr[20];        int x,y,z;        while(p--)        {            scanf("%s",opr);            if(strcmp(opr,"ADD") == 0)            {                scanf("%d%d%d",&x,&y,&z);                Add(x,y,z);            }            else if(strcmp(opr,"INSERT") == 0)            {                scanf("%d%d",&x,&y);                Insert(x,y);            }            else if(strcmp(opr,"DELETE") == 0)            {                scanf("%d",&x);                Delete(x);            }            else if(strcmp(opr,"MIN") == 0)            {                scanf("%d%d",&x,&y);                printf("%d\n",MIN(x,y));//                for(int i =1;i <= 5;i++)//                    printf("%d\n",Min[i]);            }            else if(strcmp(opr,"REVERSE") == 0)            {                scanf("%d%d",&x,&y);                Reverse(x,y);            }            else if(strcmp(opr,"REVOLVE") == 0)            {                scanf("%d%d%d",&x,&y,&z);                int t = (y-x+1);                z = (z%t+t)%t;                Revovle(x,y,z);            }        }    }    return 0;}

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