poj 3264(RMQ)

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#integer #query #each #output #input #算法

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Farmer John组织了一场奶牛终极飞盘比赛，为了确保所有奶牛都能玩得开心，他需要确定每组奶牛中最高和最矮奶牛之间的高度差。通过使用ST算法解决区间最大最小值查询问题。

Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 19690		Accepted: 9084
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

题目： http://poj.org/problem?id=3264

分析：可以算是RMQ的模板题吧，用ST算法来搞。。。

之前写了个代码一直wa，后来又写一个，几乎一字不差，居然过了。。。浪费时间。。。那个代码至今不知道哪错阿～～～

代码：

#include<iostream>
#include<cstdio>
using namespace std;
const int mm=50005;
int f[mm][16],g[mm][16];
int i,j,n,q;
void DP()
{
    int i,j,k;
    for(j=1;(1<<j)<=n;++j)
        for(i=1;i+(1<<j)-1<=n;++i)
        {
            k=i+(1<<(j-1));
            f[i][j]=max(f[i][j-1],f[k][j-1]);
            g[i][j]=min(g[i][j-1],g[k][j-1]);
        }
}
int Query(int l,int r)
{
    int m=0;
    while(l+(1<<m)<r-(1<<m)+1)++m;
    r=r-(1<<m)+1;
    return max(f[l][m],f[r][m])-min(g[l][m],g[r][m]);
}
int main()
{
    scanf("%d%d",&n,&q);
    for(i=1;i<=n;++i)scanf("%d",&f[i][0]),g[i][0]=f[i][0];
    DP();
    while(q--)scanf("%d%d",&i,&j),printf("%d\n",Query(i,j));
    return 0;
}