POJ 3264-balanced LINEUP-RMQ Problems

Bare RMQ Problem

1#include <cstdio>2#include <algorithm>3#include <cstring>4 5 using namespacestd;6 7 Const intMAXN =50010;8 9 intmin_dp[maxn][ -],max_dp[maxn][ -];Ten intMIN_MM[MAXN],MAX_MM[MAXN]; One intB[MAXN]; A intn,q; -  - voidMIN_INITRMQ (intNintb[]) the { -min_mm[0] = -1; -          for(intI=1; i<=n;i++) -         { +Min_mm[i] = ((i& (i-1)) ==0) ? min_mm[i-1]+1: min_mm[i-1]; -min_dp[i][0] =B[i]; +         } A          for(intj=1; j<=min_mm[n];j++) at                  for(intI=1; i + (1&LT;&LT;J)-1<= n;i++) -Min_dp[i][j] = min (min_dp[i][j-1],min_dp[i+ (1<< (J-1))][j-1]); -  - } -  - intMIN_RMQ (intXinty) in { -         intK = min_mm[y-x+1]; to         returnMin (min_dp[x][k],min_dp[y-(1&LT;&LT;K) +1][k]); + } -  the  * voidMAX_INITRMQ (intNintb[]) $ {Panax Notoginsengmax_mm[0] = -1; -          for(intI=1; i<=n;i++) the         { +Max_mm[i] = ((i& (i-1)) ==0) ? max_mm[i-1]+1: max_mm[i-1]; Amax_dp[i][0] =B[i]; the         } +          for(intj=1; j<=max_mm[n];j++) -                  for(intI=1; i + (1&LT;&LT;J)-1<= n;i++) $MAX_DP[I][J] = max (max_dp[i][j-1],max_dp[i+ (1<< (J-1))][j-1]); $  - } -  the intMAX_RMQ (intXinty) - {Wuyi         intK = max_mm[y-x+1]; the         returnMax (max_dp[x][k],max_dp[y-(1&LT;&LT;K) +1][k]); - } Wu  - intMain () About { $         //freopen ("input.in", "R", stdin); -          while(~SCANF ("%d%d",&n,&Q)) -         { -                  for(intI=1; i<=n;i++) A                 { +scanf"%d",&b[i]); the                 } - MIN_INITRMQ (n,b); $ MAX_INITRMQ (n,b); the  the                  for(intI=0, l,r;i<q;i++) the                 { thescanf"%d%d",&l,&R); -printf"%d\n", MAX_RMQ (L,R)-MIN_RMQ (l,r)); in                         //printf ("%d\n", MAX_RMQ (L,r)); the                 } the         } About}

POJ 3264-balanced LINEUP-RMQ Problems