Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 53703		Accepted: 25237
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

Source

USACO 2007 January Silver

题目链接：http://poj.org/problem?id=3264

分析：线段树求最大值和最小值，然后最大值减去最小值即为正解！貌似这题好像有暴力写法？

下面给出AC代码：

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <string.h>
 4 using namespace std;
 5 #define maxsize 200020
 6 typedef struct
 7 {
 8     int left,right;
 9     int maxn;
10     int minn;
11 }Node;
12 int n,m;
13 int Max,Min;
14 int num[maxsize];
15 Node tree[maxsize*20];
16 inline void buildtree(int root,int left,int right)// 构建线段树
17 {
18     int mid;
19     tree[root].left=left;
20     tree[root].right=right;// 当前节点所表示的区间
21     if(left==right)// 左右区间相同，则此节点为叶子，max 应储存对应某个学生的值
22     {
23         tree[root].maxn=num[left];
24         tree[root].minn=num[left];
25         return;
26     }
27     mid=(left+right)/2;
28     //int a,b;// 递归建立左右子树，并从子树中获得最大值
29     buildtree(2*root,left,mid);
30     buildtree(2*root+1,mid+1,right);
31     tree[root].maxn=max(tree[root*2].maxn,tree[root*2+1].maxn);
32     tree[root].minn=min(tree[root*2].minn,tree[root*2+1].minn);
33 }
34 inline void find(int root,int left,int right)// 从节点 root 开始，查找 left 和 right 之间的最大值
35 {
36     int mid;
37     //if(tree[root].left>right||tree[root].right<left)// 若此区间与 root 所管理的区间无交集
38         //return;
39     if(left==tree[root].left&&tree[root].right==right)// 若此区间包含 root 所管理的区间
40     {
41         Max=max(tree[root].maxn,Max);
42         Min=min(tree[root].minn,Min);
43         return;
44     }
45     mid=(tree[root].left+tree[root].right)/2;
46     if(right<=mid)
47         find(root*2,left,right);
48     else if(left>mid)
49         find(root*2+1,left,right);
50     else
51     {
52         find(root*2,left,mid);
53         find(root*2+1,mid+1,right);
54         //tree[root].maxn=max(tree[root*2].maxn,tree[root*2+1].maxn);
55         //tree[root].minn=min(tree[root*2].minn,tree[root*2+1].minn);
56         //return;
57     }
58 }
59 
60 int main()
61 {
62     //char c;
63     int i;
64     int x,y;
65     //scanf("d%d",&n,&m);
66     while(scanf("%d%d",&n,&m)!=EOF)
67     {
68         for(i=1;i<=n;i++)
69             scanf("%d",&num[i]);
70         buildtree(1,1,n);
71         for(i=1;i<=m;i++)
72         {
73             //getchar();
74             Max=-99999999999;
75             Min= 99999999999;
76             scanf("%d%d",&x,&y);
77             //if(c=='Q')
78                 //printf("%d\n",find(1,x,y));
79             //else
80             //{
81                // num[x]=y;
82                // update(1,x,y);
83             //}
84             find(1,x,y);
85             printf("%d\n",Max-Min);
86         }
87     }
88     return 0;
89 }