csu 1110 RMQ with Shifts（线段树） 解题报告

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 56 2 4 8 5 1 4query(3,7)shift(2,4,5,7)query(1,4)shift(1,2)query(2,2)

Sample Output

146

ac代码附上：

#include <stdio.h>#define N 101010struct rec{    int l,r,v;} t[N*5];int n,m,a[N],p[30];int minn(int a,int b){    if(a<b) return a;    return b;}void build(int root, int l, int r){    t[root].l = l;    t[root].r = r;    if( l==r )    {        t[root].v = a[l];        return ;    }    int mid = (l+r)/2;    build(root*2,l,mid);    build(root*2+1,mid+1,r);    t[root].v = minn(t[root*2].v,t[root*2+1].v);    //t[root].v=t[root*2].v+t[root*2+1].v;}void update(int root, int x){    if( t[root].l == x && t[root].r == x )    {        t[root].v = a[x];        return ;    }    int mid = (t[root].l+t[root].r)/2;    if( mid < x ) update(root*2+1,x);    else update(root*2,x);    t[root].v = minn(t[root*2].v,t[root*2+1].v);}int getv(int root,int l,int r){    if( t[root].l == l && t[root].r == r )    {        return t[root].v;    }    int mid = (t[root].l+t[root].r)/2;    if( l>mid ) return getv(root*2+1,l,r);    if( r<=mid) return getv(root*2,l,r);    return minn(getv(root*2,l,mid),getv(root*2+1,mid+1,r));}int main(){    scanf("%d%d",&n,&m);    for (int i=1; i<=n; i++)    {        scanf("%d",&a[i]);    }    scanf("\n");    build(1,1,n);    while(m--)    {        char c,c2;        scanf("%c%c%c%c%c%c",&c,&c2,&c2,&c2,&c2,&c2);        if( c=='q' )        {            int ax,bx;            scanf("%d,%d)\n",&ax,&bx);            printf("%d\n",getv(1,ax,bx));        }        else        {            int j = 0;            while(scanf("%d,",&p[++j])==1);            j--;            scanf(")\n");            int tt=a[p[1]];            for ( int i=1; i<j; i++ ) a[p[i]] = a[p[i+1]];            a[p[j]] = tt;            for ( int i=1; i<=j; i++) update(1,p[i]);        }    }    return 0;}