POJ 3264 线段树模板题

之前做的那道是区间求和的，这道题是求区间最大值和最小值之差的，感觉这道题更简单。只需在插入时把每个区间的最大值最小值求出来保存在根节点上就可以啦~(^o^)/

Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 39475		Accepted: 18524
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

Sample Output

6
3
0

 1 #include<iostream>
 2 #include<cstring>
 3 #include<algorithm>
 4 #include<cstdio>
 5 using namespace std;
 6 #define INF 0x7fffffff
 7 int minv=INF,maxv=-INF;
 8 struct node
 9 {
10     int l,r;
11     int minv,maxv;
12     int mid()
13     {
14         return (l+r)/2;
15     }
16 }tree[8000010];
17 void buildtree(int root,int l,int r)
18 {
19     tree[root].l=l;
20     tree[root].r=r;
21     tree[root].minv=INF;
22     tree[root].maxv=-INF;
23     if(l!=r)
24     {
25         buildtree(root*2+1,l,(l+r)/2);
26         buildtree(root*2+2,(l+r)/2+1,r);
27     }
28 }
29 void insert(int root,int i,int v)
30 {
31     if(tree[root].l==tree[root].r)
32     {
33         tree[root].r=i;
34         tree[root].minv=tree[root].maxv=v;
35         return;
36     }
37     tree[root].minv=min(tree[root].minv,v);
38     tree[root].maxv=max(tree[root].maxv,v);
39     if(i<=tree[root].mid())
40         insert(2*root+1,i,v);
41     else
42         insert(2*root+2,i,v);
43 }
44 void query(int root,int s,int e)
45 {
46     if(tree[root].minv>minv&&tree[root].maxv<maxv)
47         return;
48     if(tree[root].l==s&&tree[root].r==e)
49     {
50         minv=min(minv,tree[root].minv);
51         maxv=max(maxv,tree[root].maxv);
52         return;
53     }
54     if(e<=tree[root].mid())
55         query(2*root+1,s,e);
56     else if(s>tree[root].mid())
57         query(2*root+2,s,e);
58     else
59     {
60         query(2*root+1,s,tree[root].mid());
61         query(2*root+2,tree[root].mid()+1,e);
62     }
63 }
64 int main()
65 {
66     int n,q,h;
67     int i,j,k;
68     scanf("%d%d",&n,&q);
69     buildtree(0,1,n);
70     for(i=1;i<=n;i++)
71     {
72         scanf("%d",&h);
73         insert(0,i,h);
74     }
75     for(i=0;i<q;i++)
76     {
77         int s,e;
78         scanf("%d%d",&s,&e);
79         minv=INF;
80         maxv=-INF;
81         query(0,s,e);
82         printf("%d
",maxv-minv);
83     }
84     return 0;
85 }

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