暑期训练狂刷系列——poj 3264 Balanced Lineup（线段树）

题目连接：

　　http://poj.org/problem?id=3264

题目大意：

　　有n个数从1开始编号，问在指定区间内，最大数与最小数的差值是多少？

解题思路：

　　在节点中存储max，min，然后查询指定区间的max、min。

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <algorithm>
 4 using namespace std;
 5 const int maxn = 200010;
 6 const int INF = 0x3f3f3f3f;
 7 struct node
 8 {
 9     int L, R;
10     int Min, Max;
11     int Mid()
12     {
13         return (L + R) / 2;
14     }
15 };
16 node tree[maxn];
17 int Min, Max;
18 
19 void build(int root, int l, int r)
20 {
21     tree[root].L = l;
22     tree[root].R = r;
23     tree[root].Min = INF;
24     tree[root].Max = -INF;
25     if (l == r)
26         return ;
27     build (2*root+1, l, tree[root].Mid());
28     build (2*root+2, tree[root].Mid()+1, r);
29 }
30 void insert (int root, int x, int s)
31 {
32     tree[root].Min = min (tree[root].Min, s);
33     tree[root].Max = max (tree[root].Max, s);
34     if (tree[root].L == tree[root].R)
35         return ;
36     if (x <= tree[root].Mid())
37         insert (2*root+1, x, s);
38     else if (x > tree[root].Mid())
39         insert (2*root+2, x, s);
40 }
41 void query (int root, int s, int e)
42 {
43     if (tree[root].L == s && tree[root].R == e)
44     {
45         Min = min (Min, tree[root].Min);
46         Max = max (Max, tree[root].Max);
47         return ;
48     }
49     if (e <= tree[root].Mid())
50         query (2*root+1, s, e);
51     else if (s > tree[root].Mid())
52         query (2*root+2, s, e);
53     else
54     {
55         query (2*root+1, s, tree[root].Mid());
56         query (2*root+2, tree[root].Mid()+1, e);
57     }
58 }
59 int main ()
60 {
61     int n, q, num;
62     while (scanf ("%d %d", &n, &q) != EOF)
63     {
64         build(0, 1, n);
65         for (int i=1; i<=n; i++)
66         {
67             scanf ("%d", &num);
68             insert (0, i, num);
69         }
70         while (q --)
71         {
72             int s, e;
73             scanf ("%d %d", &s, &e);
74             Min = INF;
75             Max = -INF;
76             query (0, s, e);
77             printf ("%d\n", Max - Min);
78         }
79     }
80     return 0;
81 }