Problem 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
思路：还以为是把shift的索引所对的值降序排序然后更新，原来却是shift的全部索引向左移一位，首位就放到最后。直接线段树单点更新。
#include <cstdio>
#include <algorithm>
using namespace std;
int n,q,node[300000],ans,sortarr[30],num[100001],val[30];
void build(int left,int right,int index)
{
if(left==right)
{
node[index]=num[left];
}
else
{
int mid=(left+right)>>1;
build(left,mid,index<<1);
build(mid+1,right,index<<1|1);
node[index]=min(node[index<<1],node[index<<1|1]);
}
}
void update(int p,int left,int right,int index,int val)
{
int mid=(left+right)>>1;
if(left==right) node[index]=val;
else
{
if(p<=mid) update(p,left,mid,index<<1,val);
else update(p,mid+1,right,index<<1|1,val);
node[index]=min(node[index<<1],node[index<<1|1]);
}
}
void query(int L,int R,int left,int right,int index)
{
if(L<=left && right<=R)
{
ans=min(ans,node[index]);
}
else
{
int mid=(left+right)>>1;
if(L<=mid) query(L,R,left,mid,index<<1);
if(R>mid) query(L,R,mid+1,right,index<<1|1);
}
}
int main()
{
int i,j,a,b,cnt,temp;
char s[81];
while(~scanf("%d%d",&n,&q))
{
for(i=1;i<=n;i++) scanf("%d",&num[i]);
build(1,n,1);
while(q--)
{
scanf("%s",s);
if(s[0]==‘q‘)
{
sscanf(s,"query(%d,%d)",&a,&b);
ans=999999999;
query(a,b,1,n,1);
printf("%d\n",ans);
}
else
{
cnt=0;
int t=0;
for(i=0;s[i];i++) if(s[i]==‘(‘) break;
for(int j=i+1;s[j];j++){
while(s[j]!=‘,‘&&s[j]!=‘)‘){
t=t*10+s[j]-‘0‘;
j++;
}
sortarr[cnt++]=t;
t=0;
}
for(i=0;i<cnt;i++) val[i]=num[sortarr[i]];
for(i=0;i<cnt;i++)
{
update(sortarr[i],1,n,1,val[i+1<cnt?i+1:0]);
num[sortarr[i]]=val[i+1<cnt?i+1:0];
}
}
}
}
}
RMQ with Shifts(线段树单点跟新),布布扣,bubuko.com
RMQ with Shifts(线段树单点跟新)
原文：http://blog.csdn.net/faithdmc/article/details/21320449
