标签：style blog http java color strong
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
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.
n100, 000, 1q250, 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.
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)
1
4
6
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 long 10 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 }
C. RMQ with Shifts,布布扣,bubuko.com
标签：style blog http java color strong
原文地址：http://www.cnblogs.com/crackpotisback/p/3841346.html
