RMQ with Shifts(线段树)

RMQ with Shifts

Time Limit:1000MS     Memory Limit:65535KB     64bit IO Format:%I64d & %I64u

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

Hint

无

//线段树的应用，不是很难，写出来还是对线段树有更深一点的了解的

  1 #include <stdio.h>
  2 #include <string.h>
  3 
  4 #define MAXN 100005
  5 
  6 struct Node
  7 {
  8     int min;
  9     int l,r;
 10 }node[3*MAXN];//节点
 11 int pos[MAXN];//记录叶节点的位置
 12 int num[MAXN];//记录最开始的数
 13 int shift[50];//记录shift里的数
 14 
 15 int Min(int a,int b)
 16 {
 17     return a<b?a:b;
 18 }
 19 
 20 int Build(int left,int right,int k)
 21 {
 22     node[k].l=left;
 23     node[k].r=right;
 24     if (left==right)//到叶节点
 25     {
 26         node[k].min=num[left];
 27         pos[left]=k;
 28         return node[k].min;
 29     }
 30     int mid=(left+right)/2;
 31 
 32     node[k].min=Min(Build(left ,mid,2*k),Build(mid+1,right,2*k+1));
 33     return node[k].min;
 34 }
 35 
 36 int Query(int left,int right,int k)
 37 {
 38     if (left==node[k].l&&right==node[k].r)
 39     {
 40         return node[k].min;
 41     }
 42     int mid=(node[k].l+node[k].r)/2;
 43     if (left>mid) return Query(left,right,2*k+1);
 44     else if (right<=mid) return Query(left,right,2*k);
 45     return Min(Query(left,mid,2*k),Query(mid+1,right,2*k+1));
 46 }
 47 
 48 void Update(int k)
 49 {
 50     k/=2;
 51     while (k!=0)
 52     {
 53         node[k].min=Min(node[2*k].min,node[2*k+1].min);
 54         k/=2;
 55     }
 56 }
 57 
 58 int Get_shift(char str[])
 59 {
 60     int i,j;
 61     int n=0;
 62     int len=strlen(str);
 63     for (i=6;i<len;i++)
 64     {
 65         int temp=0;
 66         for (j=i;str[j]!=','&&str[j]!=')';j++)
 67         {
 68             temp+=str[j]-'0';
 69             temp*=10;
 70         }
 71         temp/=10;
 72         shift[++n]=temp;
 73         i=j;
 74     }
 75     return n;
 76 }
 77 
 78 int main()
 79 {
 80     int n,m;
 81     int i,j;
 82     scanf("%d%d",&n,&m);
 83     for (i=1;i<=n;i++)
 84         scanf("%d",&num[i]);
 85 
 86     Build(1,n,1);              //递归建树
 87     char str[50];
 88     int left,right;
 89 
 90     for (i=1;i<=m;i++)
 91     {
 92         scanf("%s",&str);
 93         if (str[0]=='q')
 94         {
 95             left=0,right=0;
 96             for (j=6;str[j]!=',';j++)
 97             {
 98                 left+=str[j]-'0';
 99                 left*=10;
100             }
101             left/=10;
102             for (j++;str[j]!=')';j++)
103             {
104                 right+=str[j]-'0';
105                 right*=10;
106             }
107             right/=10;
108             printf("%d\n",Query(left,right,1));//查找区间内最小的
109         }
110         if (str[0]=='s')
111         {
112             int shift_num=Get_shift(str);//获得shift里面的数
113 
114             int temp=node[pos[shift[1]]].min;
115             for (j=2;j<=shift_num;j++)
116                 node[pos[shift[j-1]]].min=node[pos[shift[j]]].min;
117             node[pos[shift[j-1]]].min=temp;
118             
119             for (j=1;j<=shift_num;j++)
120                 Update(pos[shift[j]]);
121         }
122 
123     }
124     return 0;
125 }
View Code


 



















            

            
posted @ 
2016-08-17 21:41 
happy_codes 
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