POJ 3264 Balanced Lineup

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

代码1（RMQ）：

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <algorithm>
#include <cstdio>
using namespace std;

const int maxn = 5e4 + 10;
int N, M;
int a[maxn];
int maxx[maxn][30], minn[maxn][30];

void RMQ(int num) {
    for(int i = 1; i <= N; i ++) {
        maxx[i][0] = a[i];
        minn[i][0] = a[i];
    }

    for(int j = 1; j < 23; j ++) {
        for(int i = 1; i <= num; i ++) {
            if(i + (1 << j) - 1<= num) {
                maxx[i][j] = max(maxx[i][j - 1], maxx[i + (1 << (j - 1))][j - 1]);
                minn[i][j] = min(minn[i][j - 1], minn[i + (1 << (j - 1))][j - 1]);
            }
        }
    }
}

int query(int l, int r) {
    int k = (int)(log(r - l + 1) / log(2.0));
    int maxnum = max(maxx[l][k], maxx[r - (1 << k) + 1][k]);
    int minnum = min(minn[l][k], minn[r - (1 << k) + 1][k]);
    return maxnum - minnum;
}

int main() {
    scanf("%d%d", &N, &M);
    for(int i = 1; i <= N; i ++)
        scanf("%d", &a[i]);

    RMQ(N);

    while(M --) {
        int st, en;
        scanf("%d%d", &st, &en);
        int ans = query(st, en);
        printf("%d
", ans);
    }
    return 0;
}

View Code

代码2（线段树）：

#include <iostream>
#include <stdio.h>
#include <cstdio>
#include <algorithm>
#include <string>
using namespace std;

const int inf = 1e8 + 10;
const int maxn = 1e6 + 10;
int N, M;
int Min, Max;
int a[maxn];

struct Node{
    int l;
    int r;
    int maxx;
    int minn;
}node[maxn];

void Build(int i, int l, int r) {
    node[i].l = l;
    node[i].r = r;
    if(l == r) {
        node[i].maxx = node[i].minn = a[l];
        return ;
    }
    int mid = (l + r) / 2;
    Build(i * 2, l, mid);
    Build(i * 2 + 1, mid + 1, r);
    node[i].maxx = max(node[i * 2].maxx, node[i * 2 + 1].maxx);
    node[i].minn = min(node[i * 2].minn, node[i * 2 + 1].minn);
}

void query(int i, int l, int r) {
    if(node[i].maxx <= Max && node[i].minn >= Min) return;
    if(node[i].l == l && node[i].r == r) {
        Max = max(Max, node[i].maxx);
        Min = min(Min, node[i].minn);
        return ;
    }
    int mid = (node[i].l + node[i].r) / 2;
    if(r <= mid) query(i * 2, l, r);
    else if(l > mid) query(i * 2 + 1, l, r);
    else {
        query(i * 2, l, mid);
        query(i * 2 + 1, mid + 1, r);
    }
}

int main() {
    while(~scanf("%d%d", &N, &M)) {
        for(int i = 1; i <= N; i ++)
            scanf("%d", &a[i]);
        Build(1, 1, N);
        while(M --) {
            int st, en;
            scanf("%d%d", &st, &en);
            Min = inf, Max = -inf;
            query(1, st, en);
            printf("%d
", Max - Min);
        }
    }
    return 0;
}

View Code

　　线段树求区间最大值和最小值的差会建线段树了有 1.. 开心突然想起来今天的 Leetcode 还没写