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 Hint 题意：有一个数组a，然后有两个操作一个query (L, R)表示数组a的第L个数到第R个数的这个区间的最小值，还有一个操作就是shift(i1, i2, i3, …, ik)，表示有k个数，然后每个数表示数组a的位置，然后将这k个数向左一个，比如A={6, 2, 4, 8, 5, 1, 4}, 然后 shift(2, 4, 5, 7)操作后就变成了{6, 8, 4, 5, 4, 1, 2}，操作了shift(1,2) 后就变成了{8, 6, 4, 5, 4, 1, 2}。 分析：显然有多个询问，可以用线段树来维护，每个节点存放区间的最小值，然后每个操作的左移可以分解成多个点修改就行了。 代码：
#include<iostream> #include<string> #include<cstdio> #include<cstring> #include<vector> #include<math.h> #include<stack> #include<map> #include<queue> #include<algorithm> using namespace std; const int inf = 0x3f3f3f3f; int Min[100005<<2];//存放每个区间的最小值 int a[100005]; //a数组 int len; char order[35]; int n,m; void push_up(int rt){//向上更新 Min[rt]=min(Min[rt*2],Min[rt*2+1]); } void bulid(int rt,int l,int r){ if (r==l){ //scanf ("%d",&Min[rt]); Min[rt]=a[l]; return ; } int mid=(r+l)/2; bulid(rt*2,l,mid); bulid(rt*2+1,mid+1,r); push_up(rt); } void update(int rt,int l,int r,int p,int v){//点修改，将位置p的值改为v if (r==l){ Min[rt]=v; return; } int mid=(r+l)/2; if (p<=mid)update(rt*2,l,mid,p,v); else update(rt*2+1,mid+1,r,p,v); push_up(rt); } int query(int rt,int l,int r,int ql,int qr){//查询区间ql,qr之间的最小值 int ans=-1; int mid=(r+l)/2; if (ql<=l&&r<=qr) return Min[rt];//在区间内直接返回 if (qr<=mid)return query(rt*2,l,mid,ql,qr);//如果全部在左区间 else if (ql>mid) return query(rt*2+1,mid+1,r,ql,qr);//全部在右区间 else {//两个区间都有取最小值 ans=min(query(rt*2,l,mid,ql,qr),query(rt*2+1,mid+1,r,ql,qr)); } return ans; } void shift(){ vector<int>v;//用来存放位置 int tmp=0; int i; for (i=0;order[i];i++)if (order[i]=='(')break; for (int j=i+1;order[j];j++){ while (order[j]!=','&&order[j]!=')'){//当遇到','或者')'说明一个数字转换好了 tmp=tmp*10+order[j]-'0';//求数字 j++; } v.push_back(tmp); tmp=0; } int tt=a[v[0]];//保存第一个数的值 for (int i=0;i<v.size()-1;i++){ update(1,1,n,v[i],a[v[i+1]]);//更新第v[i]个位置改成a[v[i]] a[v[i]]=a[v[i+1]]; } update(1,1,n,v[v.size()-1],tt);//最后一个 a[v[v.size()-1]]=tt; } int main () { while (scanf ("%d%d",&n,&m)!=EOF){ for (int i=1;i<=n;i++){ scanf ("%d",&a[i]); } bulid(1,1,n); for (int i=0;i<m;i++){ scanf ("%s",order); len=strlen(order); if (order[0]=='s'){ shift(); } else { int left=0,right=0; //sscanf(order,"query(%d,%d)",&left,&right); int i; for (i=6;order[i]!=',';i++){//求左值 left=left*10+order[i]-'0'; } for (i=i+1;order[i]!=')';i++){//求右值 right=right*10+order[i]-'0'; } int ans=query(1,1,n,left,right);//求答案 printf ("%d\n",ans); } } } return 0; }