时间限制(普通/Java):1000MS/3000MS 内存限制:65536KByte
描述
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}.
输入
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.
输出
For each query, print the minimum value (rather than index) in the requested range.
样例输入
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)
样例输出
1
4
6
每一行操作语句不超过30个字符，除去操作query,shift,(,)，真正涉及到操作数据的并不多
查询时查询区间最小值，更新时进行单点更新，直接更改每一个的值
1 #include <cstdio> 2 #include <algorithm> 3 using namespace std; 4 5 const int MAXN=1e5+5; 6 int a[MAXN],st[MAXN<<2],INF=0x3f3f3f3f; 7 8 inline int read(){ 9 int x=0,f=1; 10 char ch=getchar(); 11 while(ch<'0'||ch>'9'){ 12 if(ch=='-') 13 f=-1; 14 ch=getchar(); 15 } 16 while(ch>='0'&&ch<='9'){ 17 x=(x<<1)+(x<<3)+(ch^48); 18 ch=getchar(); 19 } 20 return x*f; 21 } 22 23 void pushup(int root){ 24 st[root]=min(st[root<<1],st[root<<1|1]); 25 } 26 27 void build(int l,int r,int root){ 28 if(l==r){ 29 st[root]=a[l]; 30 return; 31 } 32 int mid=(l+r)>>1; 33 build(l,mid,root<<1); 34 build(mid+1,r,root<<1|1); 35 pushup(root); 36 } 37 38 void update(int pos,int val,int l,int r,int root){ 39 if(pos<=l&&r<=pos){ 40 st[root]=val; 41 return; 42 } 43 int mid=(l+r)>>1; 44 if(pos<=mid)update(pos,val,l,mid,root<<1); 45 else update(pos,val,mid+1,r,root<<1|1); 46 pushup(root); 47 } 48 49 int query(int L,int R,int l,int r,int root){ 50 if(L<=l&&r<=R)return st[root]; 51 int mid=(l+r)>>1,res=INF; 52 if(L<=mid)res=min(res,query(L,R,l,mid,root<<1)); 53 if(R>mid)res=min(res,query(L,R,mid+1,r,root<<1|1)); 54 pushup(root); 55 return res; 56 } 57 58 int main(){ 59 int n,q,val[35]; 60 char c,head; 61 n=read(),q=read(); 62 for(int i=1;i<=n;i++){ 63 a[i]=read(); 64 } 65 build(1,n,1); 66 while(q--){ 67 int k=0; 68 head=getchar(); 69 while(c=getchar(),c!='('); 70 while(1){ 71 scanf("%d",&val[k++]); 72 c=getchar(); 73 if(c==')')break; 74 } 75 if(head=='q'){ 76 printf("%d ",query(val[0],val[1],1,n,1)); 77 } 78 else if(head=='s'){ 79 int bef=query(val[0],val[0],1,n,1),v; 80 for(int i=0;i<k-1;i++){ 81 v=query(val[i+1],val[i+1],1,n,1); 82 update(val[i],v,1,n,1); 83 } 84 update(val[k-1],bef,1,n,1); 85 } 86 getchar(); 87 } 88 }