POJ-3264 Balanced Lineup (线段树 基本题)

Balanced Lineup

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 57981 Accepted: 27169Case 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

6 31734251 54 62 2

Sample Output

630

#include <stdio.h>#include <string.h>#include <algorithm>#include <vector>using namespace std;#define maxn 50001int c[maxn << 2][2];void add(int o, int l, int r, int id, int v){if(l == r){c[o][0] = c[o][1] = v;return;}int mid = l + r >> 1;if(id <= mid) add(o << 1, l, mid, id, v);if(id > mid) add(o << 1 | 1, mid + 1, r, id, v);c[o][0] = min(c[o << 1][0], c[o << 1 | 1][0]);c[o][1] = max(c[o << 1][1], c[o << 1 | 1][1]);}int queryMin(int o, int l, int r, int L, int R){if(l >= L && r <= R){return c[o][0];}int mid = l + r >> 1, ans = 1e9;if(mid >= L) ans = min(ans, queryMin(o << 1, l, mid, L, R));if(mid < R) ans = min(ans, queryMin(o << 1 | 1, mid + 1, r, L, R));return ans;}int queryMax(int o, int l, int r, int L, int R){if(l >= L && r <= R){return c[o][1];}int mid = l + r >> 1, ans = 0;if(mid >= L) ans = max(ans, queryMax(o << 1, l, mid, L, R));if(mid < R) ans = max(ans, queryMax(o << 1 | 1, mid + 1, r, L, R));return ans;}int main(){int n, q, x, l, r;scanf("%d %d", &n, &q);memset(c, 0, sizeof(c));for(int i = 1; i <= n; ++i){scanf("%d", &x);add(1, 1, n, i, x);}while(q--){scanf("%d %d", &l, &r);printf("%d\n", queryMax(1, 1, n, l, r) - queryMin(1, 1, n, l, r));}}/*题意：50000个数，2e5次询问，每次询问区间最大值和最小值的差。思路：线段树裸题。*/