POJ 3264 Balanced Lineup (st or line segment tree), pojlineup

POJ 3264 Balanced Lineup (st or line segment tree), pojlineup

A-Balanced Lineup Time Limit:5000 MS Memory Limit:65536KB 64bit IO Format:% I64d & % I64uSubmit Status Practice POJ 3264 Appoint description: System Crawler)

Description

For the daily milking, Farmer John'sNCows (1 ≤N≤ 50,000) always line up in the same order. one day Farmer John decides to organize a game of Ultimate Frisbee with some of the cows. to keep things simple, he will take a contiguous range of cows from the milking lineup to play the game. however, for all the cows to have fun they shocould not differ too much in height.

Farmer John has made a listQ(1 ≤Q≤ 200,000) potential groups of cows and their heights (1 ≤Height≤ 1,000,000). For each group, he wants your help to determine the difference in height between the shortest and the tallest cow in the group.

Input

Line 1: Two space-separated integers, NAnd Q.
Lines 2 .. N+ 1: Line I+ 1 contains a single integer that is the height of cow I
Lines N+ 2 .. N+ Q+ 1: Two integers AAnd B(1 ≤ A≤ B≤ N), Representing the range of cows from ATo BIntrusive.

Output

Lines 1 .. Q: Each line contains a single integer that is a response to a reply and indicates the difference in height between the tallest and shortest cow in the range.

Sample Input

6 31734251 54 62 2

Sample Output

630

St Algorithm

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>using namespace std;#define maxn 50005int n,q;int dp[maxn][30],d[maxn][30],a[maxn];void rmq(){   for(int i=1;i<=n;i++)dp[i][0]=a[i],d[i][0]=a[i];   for(int j=1;(1<<j)<=n;j++){    for(int i=1;i+(1<<j)-1<=n;i++){        dp[i][j]=min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);        d[i][j]=max(d[i][j-1],d[i+(1<<(j-1))][j-1]);    }   }}int query1(int l,int r){    int k=0;    while((1<<(k+1))<=r-l+1)k++;    return min(dp[l][k],dp[r-(1<<k)+1][k]);}int query2(int l,int r){    int k=0;    //int k=floor(log(R-L+1.0)/log(2.0));    while((1<<(k+1))<=r-l+1)k++;    return max(d[l][k],d[r-(1<<k)+1][k]);}int main(){    int u,v;    //freopen("in.txt","r",stdin);    while(~scanf("%d%d",&n,&q)){        for(int i=1;i<=n;i++){            scanf("%d",&a[i]);        }        rmq();        for(int i=0;i<q;i++){            scanf("%d%d",&u,&v);            printf("%d\n",query2(u,v)-query1(u,v));        }    }}

Line Segment tree

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>using namespace std;#define maxn 50005#define inf 0x3f3f3f3fint n,q;int a[maxn],ll[maxn<<1],rr[maxn<<1],mi[maxn<<1],ma[maxn<<1];inline void pushup(int i){    ma[i]=max(ma[i<<1],ma[i<<1|1]);    mi[i]=min(mi[i<<1],mi[i<<1|1]);}void build(int l,int r,int i){    ll[i]=l;    rr[i]=r;    if(l==r){        mi[i]=a[l];        ma[i]=a[l];        return ;    }    int m=(ll[i]+rr[i])>>1,ls=i<<1,rs=ls|1;    build(l,m,ls);    build(m+1,r,rs);    pushup(i);}int query1(int l,int r,int i){    if(l<=ll[i]&&rr[i]<=r){        return mi[i];    }    int m=(ll[i]+rr[i])>>1,ls=i<<1,rs=ls|1;    int ans=inf;    if(l<=m)ans=min(ans,query1(l,r,ls));    if(m<r)ans=min(ans,query1(l,r,rs));    return ans;}int query2(int l,int r,int i){    if(l<=ll[i]&&rr[i]<=r){        return ma[i];    }    int m=(ll[i]+rr[i])>>1,ls=i<<1,rs=ls|1;    int ans=0;    if(l<=m)ans=max(ans,query2(l,r,ls));    if(m<r)ans=max(ans,query2(l,r,rs));    return ans;}int main(){    int u,v;    //freopen("in.txt","r",stdin);    while(~scanf("%d%d",&n,&q)){        for(int i=1;i<=n;i++){            scanf("%d",&a[i]);        }        build(1,n,1);        for(int i=0;i<q;i++){            scanf("%d%d",&u,&v);            printf("%d\n",query2(u,v,1)-query1(u,v,1));        }    }}

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