Algorithm Note--st table

Overview: An off-line algorithm that uses multiplication method to calculate interval maximum, O (nlogn) preprocessing, O (1) access.

Pretreatment:

Status: ST[I][J]:[I,I+2^J) The maximum value between

State transfer: If J equals 0,st[i][j]=0

If J is greater than 0,st[i][j]=max (st[i][j-1],st[i+2^ (j-1)][j-1]) or St[i][j]=min (st[i][j-1],st[i+2^ (j-1)][j-1])

Access:

Find the maximum value in the [L,r] Interval

K=floor (log (r-l+1))

Ans=max (St[l][k],st[r-2^k+1][k]) or Ans=min (St[l][k],st[r-2^k+1][k])

Template:

intst[n][ -],a[n];voidInit_st (intN) {     for(inti=n;i>=1; i--) {st[i][0]=A[i];  for(intj=0; i+ (1<<j-1)-1<=n;j++) {St[i][j]=min (st[i][j-1],st[i+ (1<<j-1)][j-1]); }    }}intQueryintLintR) {    intK=floor (log (r-l+1)/log (2)); returnMin (st[l][k],st[r-(1&LT;&LT;K) +1][k]);}

Example:POJ-3264

Idea: Finding the difference between the interval's value

Code:

#include <iostream>#include<cstdio>#include<string>#include<cmath>#include<algorithm>using namespacestd;#definell Long Long#definePB Push_back#defineMem (A, B) memset (A,b,sizeof (a))Const intn=1e5+5;intss[n][ -],st[n][ -],a[n];voidInit_st (intN) {     for(inti=n;i>=1; i--) {st[i][0]=A[i]; ss[i][0]=A[i];  for(intj=1; i+ (1<<j-1)-1<=n;j++) {St[i][j]=min (st[i][j-1],st[i+ (1<<j-1)][j-1]); SS[I][J]=max (ss[i][j-1],ss[i+ (1<<j-1)][j-1]); }    }}intQueryintLintR) {    intK=floor (log (r-l+1)/log (2)); returnMax (ss[l][k],ss[r-(1&LT;&LT;K) +1][k])-min (st[l][k],st[r-(1&LT;&LT;K) +1][k]);}intMain () {Ios::sync_with_stdio (false); Cin.tie (0); intN,q,l,r; CIN>>n>>Q;  for(intI=1; i<=n;i++) cin>>A[i];    Init_st (n);  while(q--) {cin>>l>>R; cout<<query (l,r) <<Endl; }    return 0;}

View Code

Reference: https://www.cnblogs.com/AireenYe/p/6270518.html

Algorithm Note--st table