poj 3264(模板RMQ)

最新推荐文章于 2018-07-31 16:35:27 发布

my_acm

最新推荐文章于 2018-07-31 16:35:27 发布

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分类专栏： RMQ

本文链接：https://blog.youkuaiyun.com/my_acm/article/details/38656287

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Farmer John在组织游戏时，需要确保每组奶牛之间的身高差距不大，以确保所有奶牛都能享受游戏的乐趣。通过输入奶牛数量、游戏组数及每组奶牛的身高，读者将学习如何计算每组中最高与最低奶牛之间的身高差。

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Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 34488		Accepted: 16203
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

Sample Output

6
3
0

Source

USACO 2007 January Silver

就是套模板啦，算法训练指南第198页有详解。

AC代码：

#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
int n,q;
int a[50010];
int d1[50010][110],d2[50010][110];
void RMQ_init(){
    for(int i=1;i<=n;i++)
        d1[i][0]=d2[i][0]=a[i];

    for(int j=1;(1<<j)<=n;j++)
    for(int i=1;i+(1<<j)-1<=n;i++){
        d1[i][j]=max(d1[i][j-1],d1[i+(1<<(j-1))][j-1]);
        d2[i][j]=min(d2[i][j-1],d2[i+(1<<(j-1))][j-1]);
    }
}
int RMQ1(int x,int y){
    int k=0;
    while((1<<(k+1)<=y-x+1)) k++;
    return max(d1[x][k],d1[y-(1<<k)+1][k]);
}
int RMQ2(int x,int y){
    int k=0;
    while(((1<<(k+1))<=y-x+1)) k++;
    return min(d2[x][k],d2[y-(1<<k)+1][k]);
}
int main(){
    while(scanf("%d%d",&n,&q)!=EOF){
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        RMQ_init();
        while(q--){
            int x,y;
            scanf("%d%d",&x,&y);
            printf("%d\n",RMQ1(x,y)-RMQ2(x,y));
        }
    }
    return 0;
}