POJ3264 Balanced Lineup

Exercises

Can be used as a template problem for the ST algorithm in RMQ

Code:

#include <iostream>#include<cstring>#include<cstdio>#include<algorithm>#include<cmath>#include<map>#include<Set>#include<vector>using namespacestd;using namespacestd;#definePB Push_back#defineMP Make_pair#defineSe Second#defineFS First#definell Long Long#defineMS (x, y) memset (x,y,sizeof (×))#defineMC (x, y) memcpy (x,y,sizeof (×))#defineLS o<<1#defineRS O<<1|1#defineSZ (x) ((int) (x). Size ())#definefor (It,c) for (__typeof ((c). Begin ()) it= (c). Begin (); it!= (c). end (); it++)typedef pair<int,int>P;Const Doubleeps=1e-9;Const intmaxn=50100;Const intn=1e9;Const intmod=1e9+7; ll read () {ll x=0, f=1;CharCh=GetChar ();  while(ch<'0'|| Ch>'9'){if(ch=='-') f=-1; ch=GetChar ();}  while(ch>='0'&&ch<='9') {x=x*Ten+ch-'0'; ch=GetChar ();} returnx*F;}//-----------------------------------------------------------------------------intW[MAXN];intN,q,l,r;intmi[maxn][ -],mx[maxn][ -];//Mi[j][i], which starts with J and is the minimum length of the 2^i, MXvoidRmq_init () {//Preprocess to the beginning of each number: The range is 2^0,2^1 ... the extremum in the ...     for(intI=1; i<=n;i++) mi[i][0]=mx[i][0]=W[i]; intM= (int) (Log (n1.0)/log (2.0));//m represents the maximum value of I in the 2^i     for(intI=1; i<=m;i++)     for(intj=1; j<=n;j++) {Mi[j][i]=mi[j][i-1];//Mi[j][i] The minimum value can be from mi[j][i-1] and mi[j+ (1<< (i-1)][i-1], is divided into half        if(J+ (1<< (I-1)) <=n) Mi[j][i]=min (mi[j][i],mi[j+ (1<< (I-1))][i-1]); Mx[j][i]=mx[j][i-1]; if(J+ (1<< (I-1)) <=n) Mx[j][i]=max (mx[j][i],mx[j+ (1<< (I-1))][i-1]); }}//the extremum of [l,r] can be calculated from [L,m],[m,r] two ranges//as long as you meet l+2^k>=r-2^k+1//2^ (k+1) >=r-l+1------>k>=log2 (r-l+1)-1;//1. When log2 (r-l+1) is an integer, K takes log (r-l+1)-1, can also take log (r-l+1), that is, the entire range//2. When log2 (r-l+1) is not an integer, K takes the upward-rounding of log (r-l+1)-1, which is the downward rounding of log (r-l+1)//synthesis 1, 2, go directly to log2 (r-l+1) to the downward rounding, that is (int) (LOG2 (r-l+1))intRmq_max (intLintR) {    intK= (int) (Log (r-l+1)*1.0)/log (2.0)); returnMax (mx[l][k],mx[r-(1&LT;&LT;K) +1][k]);//Add 1 here because, for example [1,8], with a range of 8, the 8-8+1}intRmq_min (intLintR) {    intK= (int) (Log (r-l+1)*1.0/log (2.0)); returnMin (mi[l][k],mi[r-(1&LT;&LT;K) +1][k]);}intMain () {scanf ("%d%d",&n,&q);  for(intI=1; i<=n;i++) scanf ("%d",&W[i]);    Rmq_init ();  for(intI=1; i<=q;i++) {scanf ("%d%d",&l,&R); printf ("%d\n", Rmq_max (L,R)-rmq_min (l,r)); }    return 0;}

POJ3264 Balanced Lineup