poj3264 Balanced Lineup

题意：

给定Q (1 ≤ Q ≤ 200,000)个数A1,A2 … AQ,，多次求任一区间Ai – Aj中最大数和最小数的差。

思路：

线段树。

实现：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 using namespace std;
 5
 6 const int MAXN = 65536;
 7 const int INF = 0x3f3f3f3f;
 8
 9 struct node
10 {
11     int maxn, minn;
12 };
13
14 int a[MAXN + 1];
15 node data[2 * MAXN - 1];
16 int n, q, x, y;
17
18 void init(int n_)
19 {
20     n = 1;
21     while (n < n_)
22     {
23         n *= 2;
24     }
25     for (int i = 0; i < 2 * n - 1; i++)
26     {
27         data[i].maxn = -INF;
28         data[i].minn = INF;
29     }
30 }
31
32 void update(int k, int a)
33 {
34     k += n - 1;
35     data[k].minn = data[k].maxn = a;
36     while (k)
37     {
38         k = (k - 1) / 2;
39         data[k].minn = min(data[2 * k + 1].minn, data[2 * k + 2].minn);
40         data[k].maxn = max(data[2 * k + 1].maxn, data[2 * k + 2].maxn);
41     }
42 }
43
44 int query(int a, int b, int k, int l, int r, int type)
45 {
46     if (r <= a || l >= b)
47     {
48         if (type)
49             return INF;
50         else
51             return -INF;
52     }
53     if (l >= a && r <= b)
54     {
55         if (type)
56             return data[k].minn;
57         else
58             return data[k].maxn;
59     }
60     int x = query(a, b, 2 * k + 1, l, (l + r) / 2, type);
61     int y = query(a, b, 2 * k + 2, (l + r) / 2, r, type);
62     if (type)
63         return min(x, y);
64     return max(x, y);
65 }
66
67 int main()
68 {
69     scanf("%d %d", &n, &q);
70     int n_ = n;
71     init(n);
72     for (int i = 0; i < n_; i++)
73     {
74         scanf("%d", &a[i]);
75         update(i, a[i]);
76     }
77     for (int i = 0; i < q; i++)
78     {
79         scanf("%d %d", &x, &y);
80         int s = query(x - 1, y, 0, 0, n, 1);
81         int t = query(x - 1, y, 0, 0, n, 0);
82         printf("%d\n", t - s);
83     }
84     return 0;
85 }