| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 39060 | Accepted: 18299 | |
| 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
Output
Sample Input
6 3 1 7 3 4 2 5 1 5 4 6 2 2
Sample Output
6 3 0
Source
简单线段树，维护一个区间最大值和最小值即可
AC代码：
#include <map>
#include <set>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <cstdio>
#include <cctype>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define INF 0x7fffffff
using namespace std;
const int maxn = 50005;
int n, q;
int Max[maxn << 2];
int Min[maxn << 2];
void build(int l, int r, int rt) {
if(l == r) {
scanf("%d", &Max[rt]);
Min[rt] = Max[rt];
return;
}
int mid = (l + r) >> 1;
build(l, mid, rt << 1);
build(mid + 1, r, rt << 1 | 1);
Max[rt] = max(Max[rt << 1], Max[rt << 1 | 1]);
Min[rt] = min(Min[rt << 1], Min[rt << 1 | 1]);
}
int query1(int L, int R, int l, int r, int rt) {
if(L <= l && r <= R) {
return Max[rt];
}
int ret = 0;
int mid = (l + r) >> 1;
if(L <= mid) ret = max(ret, query1(L, R, l, mid, rt << 1));
if(R >= mid + 1) ret = max(ret, query1(L, R, mid + 1, r, rt << 1 | 1));
return ret;
}
int query2(int L, int R, int l, int r, int rt) {
if(L <= l && r <= R) {
return Min[rt];
}
int ret = INF;
int mid = (l + r) >> 1;
if(L <= mid) ret = min(ret, query2(L, R, l, mid, rt << 1));
if(R >= mid + 1) ret = min(ret, query2(L, R, mid + 1, r, rt << 1 | 1));
return ret;
}
int main() {
while(scanf("%d %d", &n, &q) != EOF) {
build(1, n, 1);
for(int i = 0; i < q; i ++) {
int a, b;
scanf("%d %d", &a, &b);
int ma = query1(a, b, 1, n, 1);
int mi = query2(a, b, 1, n, 1);
printf("%d\n", ma - mi);
}
}
return 0;
}
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POJ - 3264 - Balanced Lineup （线段树）
原文地址：http://blog.csdn.net/u014355480/article/details/47080351
