本文共 2443 字，大约阅读时间需要 8 分钟。
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 54929 | Accepted: 25726 | |
| 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 31734251 54 62 2
Sample Output
630题意：给n个数字，q次询问，每次询问一个区间[a, b]内的最大值与最小值之差。
解析：很裸的一个线段树，而且不需要建好树之后都不用更新，只需查询。维护区间内的最大值和最小值即可。
代码：
#include#include #include using namespace std;#define N 50001struct node { int l, r; int Max, Min;};node a[N*4];void push_up(int i){ a[i].Max = max(a[i<<1].Max, a[i<<1|1].Max); a[i].Min = min(a[i<<1].Min, a[i<<1|1].Min);}void build(int l, int r, int i){ a[i].l = l; a[i].r = r; if(l == r){ int x; scanf("%d", &x); a[i].Max = a[i].Min = x; return; } int mid = (l + r) >> 1; build(l, mid, i<<1); build(mid+1, r, i<<1|1); push_up(i);}void query(int l, int r, int i, int &Max, int &Min){ if(a[i].l == l && a[i].r == r){ Max = a[i].Max; Min = a[i].Min; return; } int mid = (a[i].l + a[i].r)>>1; if(l > mid) query(l, r, i<<1|1, Max, Min); else if(r <= mid) query(l, r, i<<1, Max, Min); else{ int Max0, Max1, Min0, Min1; query(l, mid, i<<1, Max0, Min0); query(mid+1, r, i<<1|1, Max1, Min1); Max = max(Max0, Max1); Min = min(Min0, Min1); }}int main(){ int n, q, a, b, Max, Min; while(~scanf("%d%d", &n, &q)){ build(1, n, 1); for(int i = 0; i < q; i++){ scanf("%d%d", &a, &b); if(a == b){ puts("0"); continue; } query(a, b, 1, Max, Min); printf("%d\n", Max-Min); } } return 0;}
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