POJ Balanced Lineup 3264

                                                             

Balanced Lineup

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 48372 Accepted: 22684Case Time Limit: 2000MS

Description

For the daily milking, Farmer John's N cows (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 should not differ too much in height.

Farmer John has made a list of Q (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, N andQ.
Lines 2..N+1: Line i+1 contains a single integer that is the height of cowi
Lines N+2..N+Q+1: Two integers A and B (1 ≤A ≤B ≤ N), representing the range of cows from A toB inclusive.

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

Source

USACO 2007 January Silver

题意：给定区间，求区间最大最小值的差

分析：可以用线段树做，没有更新

　　　也可以用RMQ，代码也不多

//线段树#include <iostream>#include <algorithm>#include <stdio.h>#include <stdlib.h>using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define maxx 51000#define inf 1123456long long Max[maxx<<2];long long Min[maxx<<2];void gengxin(int rt){    Max[rt]=max(Max[rt<<1],Max[rt<<1|1]);    Min[rt]=min(Min[rt<<1],Min[rt<<1|1]);}void build(int l,int r,int rt){    if(l==r)    {        scanf("%lld",&Max[rt]);        Min[rt]=Max[rt];                return ;    }    int m=(l+r)>>1;    build(lson);    build(rson);    gengxin(rt);}long long querymax(int x,int y,int l,int r,int rt){    if(x<=l&&r<=y)    {        return Max[rt];       // printf("%lld\n",Max[rt]);    }    int m=(l+r)>>1;    long long ret=-inf;    if(x<=m)         ret=max(ret,querymax(x,y,lson));    if(m<y)         ret=max(ret,querymax(x,y,rson));    return ret;}long long querymin(int x,int y,int l,int r,int rt){    if(x<=l&&r<=y)    {        return Min[rt];    }    int m=(l+r)>>1;    long long ret=inf;    if(x<=m)         ret=min(ret,querymin(x,y,lson));    if(m<y)         ret=min(ret,querymin(x,y,rson));    return ret;}int main(){    int n,m,i,j,x,y;    while(~scanf("%d%d",&n,&m))    {    build(1,n,1);    while(m--)    {        scanf("%d%d",&x,&y);        long long l=querymax(x,y,1,n,1);        long long s=querymin(x,y,1,n,1);;        printf("%lld\n",l-s);    }    }    return 0;}

//RMQ#include <iostream>#include <stdio.h>#include <algorithm>#include <math.h>using namespace std;//F[i][j]代表【i,i+(1<<j)+1】区间中的最值int Fmax[51234][20];int Fmin[51234][20];void f(int n){    int i,j;    for(j=1;j!=20;j++)    {        for(i=1;i<=n;i++)        {            if(i+(1<<j)-1<=n)            {                //所以是j-1，前面的要加上                Fmax[i][j]=max(Fmax[i][j-1],Fmax[i+(1<<(j-1))][j-1]);                Fmin[i][j]=min(Fmin[i][j-1],Fmin[i+(1<<(j-1))][j-1]);            }        }    }}int main(){    int n,m,i,j;    int a,b;    scanf("%d%d",&n,&m);    for(i=1;i<=n;i++)    {        scanf("%d",&Fmax[i][0]);        Fmin[i][0]=Fmax[i][0];    }    f(n);    while(m--)    {        scanf("%d%d",&a,&b);        int k;        k=int(log(b-a+1.0)/log(2.0));        int maxx=max(Fmax[a][k],Fmax[b-(1<<k)+1][k]);        int minn=min(Fmin[a][k],Fmin[b-(1<<k)+1][k]);        printf("%d\n",maxx-minn);    }    return 0;}