poj 3246 Balanced Lineup

最新推荐文章于 2020-07-05 18:38:53 发布

原创最新推荐文章于 2020-07-05 18:38:53 发布 · 198 阅读

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本文介绍了一种解决农民约翰组织奶牛游戏时确保高度差异最小化的算法。使用RMQ(st)算法预先处理奶牛高度数据，快速计算任意指定区间内最高与最矮奶牛之间的高度差。

Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 59869		Accepted: 28040
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

题意：有n头牛每头牛有自己特定的高度，找出对于区间中最高的牛和最低的牛高度之差

用RMQ st算法进行求解

//st算法  rmq

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
const int maxn=50005;
int a[maxn],dp[maxn][30],dp1[maxn][30],vis[maxn];//dp[i][j]表示从第i位开始连续的2^j个数中的最小值
void init(int n)
    {
        for (int i=1;i<=n;i++)
            {
                dp[i][0]=a[i],dp1[i][0]=a[i];
            }
        for (int j=1;(1<<j)<=n;j++)
            for (int i=1;i+(1<<j)-1<=n;i++)
                {
                    dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
                    dp1[i][j]=min(dp1[i][j-1],dp1[i+(1<<(j-1))][j-1]);
                }
    }
int minrmq(int l,int r){
    int k=log2(r-l+1);
    return min(dp1[l][k],dp1[r-(1<<k)+1][k]);
}
int maxrmq(int l,int r){
    int k=log2(r-l+1);
    return max(dp[l][k],dp[r-(1<<k)+1][k]);
}
int main()
    {
        int n,m;
        int A,B;
        while (scanf("%d%d",&n,&m)!=EOF)
            {
                for (int i=1;i<=n;i++)
                    scanf("%d",&a[i]);
                init(n);
                for (int i=0;i<m;i++)
                    {
                        scanf("%d%d",&A,&B);
                        printf("%d\n",maxrmq(A,B)-minrmq(A,B));
                    }
            }
        return 0;
    }

还有一种线段数的求法，暂时先放一下，下次更新