RMQ with Shifts

RMQ with Shifts

Time Limit: 1000MS Memory Limit: 65535KB 64bit IO Format:

Submit Status

Description

In the traditional RMQ (Range Minimum Query) problem, we have a static array A. Then for each query (L, R) (L<=R), we report the minimum value among A[L], A[L+1], …, A[R]. Note that the indices start from 1, i.e. the left-most element is A[1].

In this problem, the array A is no longer static: we need to support another operation shift(i1, i2, i3, …, ik) (i1<i2<...<ik, k>1): we do a left “circular shift” of A[i1], A[i2], …, A[ik]. 

For example, if A={6, 2, 4, 8, 5, 1, 4}, then shift(2, 4, 5, 7) yields {6, 8, 4, 5, 4, 1, 2}. After that, shift(1,2) yields{8, 6, 4, 5, 4, 1, 2}.

Input

There will be only one test case, beginning with two integers n, q (1<=n<=100,000, 1<=q<=120,000), the number of integers in array A, and the number of operations. The next line contains n positive integers not greater than 100,000, the initial elements in array A. Each of the next q lines contains an operation. Each operation is formatted as a string having no more than 30 characters, with no space characters inside. All operations are guaranteed to be valid. Warning: The dataset is large, better to use faster I/O methods.

Output

For each query, print the minimum value (rather than index) in the requested range.

Sample Input

7 5 6 2 4 8 5 1 4 query(3,7) shift(2,4,5,7) query(1,4) shift(1,2) query(2,2)

Sample Output

1 4 6

线段树模板题

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int a[100010];struct node{    int l, r, mmin;}t[100010*4];void build(int i, int l, int r){    t[i].l = l;    t[i].r = r;    if(l==r){        t[i].mmin = a[l];        return ;    }    int mid = (l+r)/2;    build(i*2, l, mid);    build(i*2+1, mid+1, r);    t[i].mmin = min(t[i*2].mmin, t[i*2+1].mmin);}int query(int i, int x, int y){    int l = t[i].l;    int r = t[i].r;    if(x==l&&y==r)        return t[i].mmin;    int mid = (l+r)/2;    if(x>mid)        return query(i*2+1, x, y);    else if(y<=mid)        return query(i*2, x, y);    else        return min(query(i*2, x, mid),query(i*2+1, mid+1, y));}void update(int i, int x, int y){    int l = t[i].l;    int r = t[i].r;    if(l==x&&l==r){        t[i].mmin = y;        a[x] = y;        return ;    }    int mid = (l+r)/2;    if(x <= mid)        update(i*2, x, y);    else        update(i*2+1, x, y);    t[i].mmin = min(t[i*2].mmin, t[i*2+1].mmin);}int main(){    int n, m, i, j, k, x, y, p, o;    char ch[110];    int b[110];    while(~scanf("%d%d",&n,&m)){        for(i=1; i<=n; i++)            scanf("%d",&a[i]);           // getchar();        build(1, 1, n);        for(i=0; i<m; i++){            scanf("%s",ch);            int l = strlen(ch);            o = 1;            memset(b, 0, sizeof(b));            if(ch[0]=='s'){                for(j=0; j<l; j++){                    p = 0;                    if(ch[j]>='0'&&ch[j]<='9'){                        for(k=j; k<l; k++){                            if(ch[k]>='0'&&ch[k]<='9')                                p = p*10+(ch[k]-'0');                            else{                                j = k;                                break;                            }                        }                    }                    if(p!=0)                        b[o++] = p;                }                int P = a[b[1]];               // cout<<b[1]<<endl;                for(k=1; k<=o-1; k++){                    update(1, b[k], a[b[k+1]]);                  //  a[b[k]] = a[b[k+1]];//                  for(int pp=1; pp<=n; pp++)//                    cout<<a[pp]<<" ";//                  cout<<endl;                }                update(1, b[o-1], P);              //  a[b[o-1]] = a[P];            }            else if(ch[0]=='q'){                x = y = 0;                int h;                for(j=0; j<l; j++){                    if(ch[j]=='(')                        break;                }                for(h=j+1; h<l; h++){                    if(ch[h]>='0'&&ch[h]<='9')                        x = x*10+(ch[h]-'0');                    else                        break;                }                for(j=h+1; j<l; j++){                    if(ch[j]>='0'&&ch[j]<='9')                        y = y*10+(ch[j]-'0');                    else                        break;                }                printf("%d\n",query(1, x, y));            }           // getchar();        }    }    return 0;}