UVa 12299 RMQ with Shifts(线段树)

线段树,没了..

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#include<cstdio>

#include<cstring>

#include<iostream>

#include<algorithm>

#include<cctype>

  

#define rep(i,n) for(int i=0;i<n;i++)

#define clr(x,c) memset(x,c,sizeof(x))

#define Rep(i,l,r) for(int i=l;i<=r;i++)

  

using namespace std;

  

const int inf=0x7fffffff;

const int maxn=100005;

  

int a[maxn];

int minv[1<<18];

  

void build(int o,int l,int r) {

int m=l+(r-l)/2;

if(l==r) minv[o]=a[l];

else {

build(o*2,l,m);

build(o*2+1,m+1,r);

minv[o]=min(minv[o*2],minv[o*2+1]);

}

}

  

int p,v;

void update(int o,int l,int r) {

int m=l+(r-l)/2;

if(l==r) minv[o]=v;

else {

p<=m ? update(o*2,l,m) : update(o*2+1,m+1,r);

minv[o]=min(minv[o*2],minv[o*2+1]);

}

}

  

int ql,qr;

int query(int o,int l,int r) {

int m=l+(r-l)/2,ans=inf;

if(ql<=l && qr>=r) return minv[o];

if(ql<=m) ans=min(ans,query(o*2,l,m));

if(qr>m) ans=min(ans,query(o*2+1,m+1,r));

return ans;

}

int main()

{

freopen("test.in","r",stdin);

freopen("test.out","w",stdout);

int n,q;

scanf("%d%d",&n,&q);

Rep(i,1,n) scanf("%d",&a[i]);

build(1,1,n);

int cmd[20],change[20],cnt;

char s[100];

while(q--) {

scanf("%s",s);

int len=strlen(s);

cmd[cnt=0]=0;

for(int i=6;i<len;++i)

if(isdigit(s[i])) (cmd[cnt]*=10)+=s[i]-'0';

else cmd[++cnt]=0;

if(s[0]=='q') {

ql=cmd[0],qr=cmd[1];

printf("%d\n",query(1,1,n));

} else {

rep(i,cnt) change[i]=a[cmd[i]];

rep(i,cnt) { v=a[p=cmd[i]]=change[(i+1)%cnt]; update(1,1,n); }

}

}

return 0;

}

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12299 RMQ with Shifts

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

shif t(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 shif t(2, 4, 5, 7) yields {6, 8, 4, 5, 4, 1, 2}. After that,

shif t(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 ≤ 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 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