(模板题)poj 3264 Balanced Lineup(RMQ的ST算法)

Balanced Lineup Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 49596   Accepted: 23232 Case 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 3 1 7 3 4 2 5 1 5 4 6 2 2 Sample Output 6 3 0 Source USACO 2007 January Silver

提示

题意：

给出一长度为N(1<=N<=50000)的数列，对该数列进行Q(1<=Q<=200000)次询问，求出[a,b]区间最大值与最小值的差。

思路：

ST算法基础题，了解原理套模板。

ST的讲解：http://blog.csdn.net/insistgogo/article/details/9929103

示例程序

Source Code

Problem: 3264		Code Length: 1184B
Memory: 6860K		Time: 3563MS
Language: GCC		Result: Accepted
#include 
#include 
int a[50000],f[2][50000][16];
int max(int x,int y)
{
    if(x>y)
    {
        return x;
    }
    else
    {
        return y;
    }
}
int min(int x,int y)
{
    if(xi;i++)
    {
        f[0][i][0]=a[i];
        f[1][i][0]=a[i];
    }
    for(i=1;n>=(1<i1+(1<i;i++)
    {
        scanf("%d",&a[i]);
    }
    bulid(n);
    for(i=1;q>=i;i++)
    {
        scanf("%d %d",&x,&y);
        x--;
        y--;
        printf("%d\n",rmq(x,y,0)-rmq(x,y,1));
    }
	return 0;
}