poj3580 SuperMemo (Splay+区间内向一个方向移动)

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Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 13550 Accepted: 4248Case 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

题意：给你n个数组成的一个序列，让你写一个数据结构支持下列操作：1.区间[x,y]增加c 2.区间[x,y]翻转 3.区间[x,y]向右移动c次，如1 2 3 4 5向右移动2次就变成4 5 1 2 3。4.在第x个数后插入c 5.删除第x个数 6.求出区间[x,y]内的最小值。

思路：要维护rev[],mx[],add[]分别表示旋转标记，最小值以及成段增加的标记，然后这题和其他题不用的地方在于多了一个区间内移动的操作，我们要先求出c被y-x+1除后的余数，如果为0就不变，如果不为0，那么画图可以看出这个操作等价于把[y-c+1,y]移到[x,y-c]这个区间的前面，然后就可以直接做了。

#include<iostream>#include<stdio.h>#include<stdlib.h>#include<string.h>#include<math.h>#include<vector>#include<map>#include<set>#include<string>#include<bitset>#include<algorithm>using namespace std;typedef long long ll;typedef long double ldb;#define inf 99999999#define pi acos(-1.0)#define Key_value ch[ch[rt][1]][0]#define maxn 1000010int n;int cnt,rt;int pre[maxn],ch[maxn][2],sz[maxn],rev[maxn],zhi[maxn],mx[maxn],add[maxn];int b[maxn],tot2;//内存池和容量int a[maxn];void update_rev(int x){    if(!x)return;       //!!!    rev[x]^=1;    swap(ch[x][0],ch[x][1]);}void update_add(int x,int value){    zhi[x]+=value;    add[x]+=value;    mx[x]+=value;}void pushdown(int x){    if(rev[x]){        update_rev(ch[x][0]);        update_rev(ch[x][1]);        rev[x]=0;    }    if(add[x]){        update_add(ch[x][0],add[x]);        update_add(ch[x][1],add[x]);        add[x]=0;    }}void pushup(int x){    int t=zhi[x];    if(ch[x][0])t=min(t,mx[ch[x][0] ]);    if(ch[x][1])t=min(t,mx[ch[x][1] ]);    mx[x]=t;    sz[x]=sz[ch[x][0] ]+sz[ch[x][1] ]+1;}void Treavel(int x){    if(x)    {        pushdown(x);        Treavel(ch[x][0]);        printf("结点：%2d: 左儿子 %2d 右儿子 %2d 父结点 %2d size = %2d zhi = %2d minx=%2d add=%2d\n",x,ch[x][0],ch[x][1],pre[x],sz[x],zhi[x],mx[x],add[x]);        Treavel(ch[x][1]);    }}void debug(){    printf("root:%d\n",rt);    Treavel(rt);}void newnode(int &x,int father,int value){    if(tot2)x=b[tot2--];    else x=++cnt;    pre[x]=father;ch[x][0]=ch[x][1]=0;sz[x]=1;rev[x]=0;zhi[x]=value;mx[x]=value;add[x]=0;}void build(int &x,int l,int r,int father){    if(l>r)return;    int mid=(l+r)/2;    newnode(x,father,a[mid]);    build(ch[x][0],l,mid-1,x);    build(ch[x][1],mid+1,r,x);    pushup(x);}void init(){    cnt=rt=tot2=0;    pre[0]=ch[0][0]=ch[0][1]=sz[0]=rev[0]=zhi[0]=mx[0]=add[0]=0;    newnode(rt,0,-1);    newnode(ch[rt][1],rt,-1);    build(Key_value,1,n,ch[rt][1]);    pushup(ch[rt][1]);    pushup(rt);}void rotate(int x,int p){    int y=pre[x];    pushdown(y);pushdown(x);    ch[y][!p]=ch[x][p];    pre[ch[x][p] ]=y;    if(pre[y])ch[pre[y] ][ch[pre[y] ][1]==y ]=x;    pre[x]=pre[y];    ch[x][p]=y;    pre[y]=x;    pushup(y);pushup(x);}void splay(int x,int goal){    pushdown(x);    while(pre[x]!=goal){        if(pre[pre[x] ]==goal){            pushdown(pre[x]);pushdown(x);            rotate(x,ch[pre[x]][0]==x);        }        else{            int y=pre[x];int z=pre[y];            pushdown(z);pushdown(y);pushdown(x);            int p=ch[pre[y] ][0]==y;            if(ch[y][p]==x )rotate(x,!p);            else rotate(y,p);            rotate(x,p);        }    }    if(goal==0)rt=x;    pushup(x);}int get_kth(int x,int k){    int i,j;    pushdown(x);    int t=sz[ch[x][0] ]+1;    if(t==k)return x;    if(t<k)return get_kth(ch[x][1],k-t);    return get_kth(ch[x][0],k);}void erase(int x){    if(x==0)return; //!!!    b[++tot2]=x;    erase(ch[x][0]);    erase(ch[x][1]);}void Add(int x,int y,int c){    splay(get_kth(rt,x),0);    splay(get_kth(rt,y+2),rt);    update_add(Key_value,c);    pushup(ch[rt][1]);    pushup(rt);}void Reverse(int x,int y){    splay(get_kth(rt,x),0);    splay(get_kth(rt,y+2),rt);    update_rev(Key_value);    pushup(ch[rt][1]);    pushup(rt);}void Insert(int x,int value){    int i,j;    splay(get_kth(rt,x+1),0);    splay(get_kth(rt,x+2),rt);    newnode(Key_value,ch[rt][1],value);    pushup(ch[rt][1]);    pushup(rt);}void Rotate(int x,int y,int c){    int len=y-x+1;    c=(c%len+len)%len;    if(c==0)return;    splay(get_kth(rt,y-c+1),0);    splay(get_kth(rt,y+2 ),rt);    int tmp=Key_value;    Key_value=0;    pushup(ch[rt][1]);    pushup(rt);    splay(get_kth(rt,x),0 );    splay(get_kth(rt,x+1),rt);    Key_value=tmp;    pre[tmp]=ch[rt][1];    pushup(ch[rt][1]);    pushup(rt);}void Delete(int pos,int tot){    splay(get_kth(rt,pos),0);    splay(get_kth(rt,pos+tot+1),rt);    erase(Key_value);    Key_value=0;    pushup(ch[rt][1]);    pushup(rt);}int Get_min(int x,int y){    splay(get_kth(rt,x),0);    splay(get_kth(rt,y+2),rt);    return mx[Key_value];}int main(){    int m,i,j,pos,tot,c,d,e,f;    char s[10];    while(scanf("%d",&n)!=EOF)    {        for(i=1;i<=n;i++){            scanf("%d",&a[i]);        }        scanf("%d",&m);        init();        //debug();        for(i=1;i<=m;i++){            scanf("%s",s);            if(s[0]=='A'){                scanf("%d%d%d",&c,&d,&e);                Add(c,d,e);            }            if(s[0]=='R' && s[3]=='E'){                scanf("%d%d",&c,&d);                Reverse(c,d);            }            if(s[0]=='R' && s[3]=='O'){                scanf("%d%d%d",&c,&d,&e);                Rotate(c,d,e);            }            if(s[0]=='I'){                scanf("%d%d",&c,&d);                Insert(c,d);            }            if(s[0]=='D'){                scanf("%d",&c);                Delete(c,1);            }            if(s[0]=='M'){                scanf("%d%d",&c,&d);                printf("%d\n",Get_min(c,d));                //printf("1\n");            }            //debug();        }    }    return 0;}

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