【POJ 3264】 Balanced Lineup (RMQ)

【POJ 3264】 Balanced Lineup (RMQ)

Balanced Lineup

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 40540 Accepted: 19056Case 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 and 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 A and B (1 ≤ A ≤ B ≤ N), representing the range of cows from A to B 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

RMQ裸题 求区间最值差 不过好久不写RMQ还是手生了 调了好久

C++比G++快2000MS。。。不过C++函数参数的兼容……唉……CE了发

代码如下:

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>using namespace std;int rqx[50005][18],rqn[50005][18];int n;void GetRMQ(){    for(int j = 1; (1<<j) < n; ++j)    {        for(int i = 0; i+(1<<j)-1 < n; ++i)        {            rqx[i][j] = max(rqx[i][j-1],rqx[i+(1<<(j-1))][j-1]);            rqn[i][j] = min(rqn[i][j-1],rqn[i+(1<<(j-1))][j-1]);        }    }}int main(){    int q,i,j,k,mx,a,b;    scanf("%d %d",&n,&q);    for(i = 0; i < n; ++i)    {        scanf("%d",&rqx[i][0]);        rqn[i][0] = rqx[i][0];    }    GetRMQ();    while(q--)    {        scanf("%d %d",&a,&b);        --a,--b;        k = log10(1.0*(b-a+1))/log10(1.0*2);        printf("%d\n",max(rqx[a][k],rqx[b-(1<<k)+1][k])-min(rqn[a][k],rqn[b-(1<<k)+1][k]));    }    return 0;}