Balanced Lineup Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 50004 Accepted: 23434 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
6 3 1 7 3 4 2 5 1 5 4 6 2 2 Sample Output
6 3 0 Source
USACO 2007 January Silver 用线段树查询每个区间上最大值与最小值的差值，用一个线段树维护最小值，一个维护最大值。
#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <iostream>
#include <stack>
#include <cmath>
#include <string>
#include <vector>
#include <cstdlib>
//#include <bits/stdc++.h>
//#define LOACL#define space " "
usingnamespace std;
typedef longlong LL;
typedef __int64 Int;
typedef pair<int, int> paii;
constint INF = 0x3f3f3f3f;
constdouble ESP = 1e-5;
constdouble PI = acos(-1.0);
constint MOD = 1e9 + 7;
constint MAXN = 200000 + 10;
int data[MAXN];
struct node {int l, r, value;
} segmin[MAXN], segmax[MAXN];
int build_segm_max(int x, int lson, int rson) {segmax[x].l = lson;segmax[x].r = rson;if (lson != rson) {int a = build_segm_max(x << 1, lson, (rson + lson)/2);int b = build_segm_max((x << 1) + 1, (rson + lson)/2 + 1, rson);return segmax[x].value = max(a, b);}return segmax[x].value = data[lson];
}
int build_segm_min(int x, int lson, int rson) {segmin[x].l = lson;segmin[x].r = rson;if (lson != rson) {int a = build_segm_min(x << 1, lson, (rson + lson)/2);int b = build_segm_min((x << 1) + 1, (rson + lson)/2 + 1, rson);return segmin[x].value = min(a, b);}return segmin[x].value = data[lson];
}
int query_seg_max(int x, int lson, int rson) {if (segmax[x].l > rson || segmax[x].r < lson) return0;if (segmax[x].l >= lson && segmax[x].r <= rson) return segmax[x].value;int a = query_seg_max(x << 1, lson, rson);int b = query_seg_max((x << 1) + 1, lson, rson);return max(a, b);
}
int query_seg_min(int x, int lson, int rson) {if (segmin[x].l > rson || segmin[x].r < lson) return INF;if (segmin[x].l >= lson && segmin[x].r <= rson) return segmin[x].value;int a = query_seg_min(x << 1, lson, rson);int b = query_seg_min((x << 1) + 1, lson, rson);return min(a, b);
}
int main() {int N, Q, a, b;scanf("%d%d", &N, &Q);for (int i = 1; i <= N; i++) scanf("%d", &data[i]);build_segm_max(1, 1, N); build_segm_min(1, 1, N);for (int i = 0; i < Q; i++) {scanf("%d%d", &a, &b);printf("%d\n", query_seg_max(1, a, b) - query_seg_min(1, a, b));}return0;
}
