(简单) POJ 3264 Balanced Lineup，RMQ。

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.

 

　　题目就是求RMQ，水题。

 

代码如下：

// ━━━━━━神兽出没━━━━━━
//      ┏┓       ┏┓
//     ┏┛┻━━━━━━━┛┻┓
//     ┃           ┃
//     ┃     ━     ┃
//     ████━████   ┃
//     ┃           ┃
//     ┃    ┻      ┃
//     ┃           ┃
//     ┗━┓       ┏━┛
//       ┃       ┃
//       ┃       ┃
//       ┃       ┗━━━┓
//       ┃           ┣┓
//       ┃           ┏┛
//       ┗┓┓┏━━━━━┳┓┏┛
//        ┃┫┫     ┃┫┫
//        ┗┻┛     ┗┻┛
//
// ━━━━━━感觉萌萌哒━━━━━━

// Author        : WhyWhy
// Created Time  : 2015年07月17日 星期五 16时52分31秒
// File Name     : 3264.cpp

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>

using namespace std;

const int MaxN=50004;

int dp1[MaxN][20],dp2[MaxN][20];
int logN[MaxN];

void init(int N,int num[])
{
    logN[0]=-1;

    for(int i=1;i<=N;++i)
    {
        dp1[i][0]=num[i];
        dp2[i][0]=num[i];
        logN[i]=logN[i-1]+((i&(i-1))==0);
    }

    for(int j=1;j<=logN[N];++j)
        for(int i=1;i+(1<<j)-1<=N;++i)
        {
            dp1[i][j]=max(dp1[i][j-1],dp1[i+(1<<(j-1))][j-1]);
            dp2[i][j]=min(dp2[i][j-1],dp2[i+(1<<(j-1))][j-1]);
        }
}

int RMQ(int x,int y)
{
    int k=logN[y-x+1];

    return max(dp1[x][k],dp1[y-(1<<k)+1][k])-min(dp2[x][k],dp2[y-(1<<k)+1][k]);
}

int num[MaxN];

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);

    int N,Q;
    int a,b;

    while(~scanf("%d %d",&N,&Q))
    {
        for(int i=1;i<=N;++i)
            scanf("%d",&num[i]);

        init(N,num);

        while(Q--)
        {
            scanf("%d %d",&a,&b);

            printf("%d\n",RMQ(a,b));
        }
    }
    
    return 0;
}

View Code

 

转载于:https://www.cnblogs.com/whywhy/p/4655138.html