首页 > 代码库 > C. RMQ with Shifts

C. RMQ with Shifts

2024-07-11 22:06:16   222人阅读

C. RMQ with Shifts

1000ms
1000ms
131072KB
 
64-bit integer IO format: %lld      Java class name: Main
 
 

In the traditional RMQ (Range Minimum Query) problem, we have a static array A. Then for each query (L, R) (L$ \le$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$ \le$n$ \le$100, 000, 1$ \le$q$ \le$250, 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

解题：RMQ问题，更新比较有新意。。。。。。。。。。

 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <vector> 6 #include <climits> 7 #include <algorithm> 8 #include <cmath> 9 #define LL long long10 using namespace std;11 const int maxn = 100010;12 struct node{13     int lt,rt,minVal;14 }tree[maxn<<2];15 int d[maxn],u[30],cnt;16 void build(int lt,int rt,int v){17     tree[v].lt = lt;18     tree[v].rt = rt;19     if(lt == rt){20         tree[v].minVal = d[lt];21         return;22     }23     int mid = (lt+rt)>>1;24     build(lt,mid,v<<1);25     build(mid+1,rt,v<<1|1);26     tree[v].minVal = min(tree[v<<1].minVal,tree[v<<1|1].minVal);27 }28 int query(int lt,int rt,int v){29     if(tree[v].lt == lt && tree[v].rt == rt) return tree[v].minVal;30     int mid = (tree[v].lt+tree[v].rt)>>1;31     if(rt <= mid) return query(lt,rt,v<<1);32     else if(lt > mid) return query(lt,rt,v<<1|1);33     else return min(query(lt,mid,v<<1),query(mid+1,rt,v<<1|1));34 }35 void update(int lt,int rt,int v){36     if(tree[v].lt == tree[v].rt){37         tree[v].minVal = d[tree[v].lt];38         return;39     }40     int mid = (tree[v].lt+tree[v].rt)>>1;41     if(u[rt] <= mid) update(lt,rt,v<<1);42     else if(u[lt] > mid) update(lt,rt,v<<1|1);43     else{44         int i;45         for(i = lt; u[i] <= mid; i++);46         update(lt,i-1,v<<1);47         update(i,rt,v<<1|1);48     }49     tree[v].minVal = min(tree[v<<1].minVal,tree[v<<1|1].minVal);50 }51 int main(){52     int n,m,i,j,len,temp;53     char str[100];54     while(~scanf("%d%d",&n,&m)){55         for(i = 1; i <= n; i++)56             scanf("%d",d+i);57             build(1,n,1);58         for(i = 0; i < m; i++){59             scanf("%s",str);60             len = strlen(str);61             for(cnt = j = 0; j < len;){62                 if(str[j] < ‘0‘ || str[j] > ‘9‘) {j++;continue;}63                 temp = 0;64                 while(j < len && str[j] >= ‘0‘ && str[j] <= ‘9‘) {temp = temp*10 + (str[j]-‘0‘);j++;}65                 u[cnt++] = temp;66             }67             if(str[0] == ‘q‘){68                 printf("%d\n",query(u[0],u[1],1));69             }else{70                 temp = d[u[0]];71                 for(cnt--,j = 0; j < cnt; j++)72                     d[u[j]] = d[u[j+1]];73                 d[u[j]]  = temp;74                 update(0,cnt,1);75             }76         }77     }78     return 0;79 }
View Code

 


联系
我们