poj3580 SuperMemo (Splay+区间内向一个方向移动)

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

1. 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}
2. 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}
3. 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}
4. 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}
5. DELETE x: delete A_x. For example, performing "DELETE 2" on {1, 2, 3, 4, 5} results in {1, 3, 4, 5}
6. 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

5
1
2
3
4
5
2
ADD 2 4 1
MIN 4 5

Sample Output

5

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 1000010
int 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;
}