Hdu 3603

Coach Yehr’s punishment

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 397 Accepted Submission(s): 114

Problem Description

During the Mult-University Trainging,Coach Yehr asks all the ACM teammates to jog at 6:30AM.But 6:30 is too early,there are always somebody might be late.Coach Yehr likes AC sequence very much,the AC sequence is a number sequence with all the elements different.A sequence (S₁ ,S₂ ,S₃ ……S_n ) is a AC sequence if S₁ ,S₂ ,S₃ ……S_n are all different. There are N teammates,the time(in second time) every teammate’arrival make a number sequence with length N. In order to punish the laters,Coach Yehr give them a puzzle,Coach Yehr choose a subsequence from S_a to S_b ,the laters must tell Coach Yehr the longest length of AC sequence in the subsequence as soon as possible.

Input

There are multiply text cases.You must deal with it until the end of file.
The first line of each test case is an interger N,indicates the number of ACM teammates;
The second line have N intergers,the i-th number indicates the i-th teammate’s arrival time.
The third line is an interger M indicates Coach Yehr will ask M times;
The follow M lines,each line have two intergers a and b,indicate the interval of the sequence.

Output

For each query,you have to print the longest length of AC sequence in the subsequence in a single line.

Sample Input

8 3 2 5 6 8 3 2 6 2 2 4 1 8 6 5 3 1 2 3 4 1 6 3 3

Sample Output

3 5 4 2

二分＋ＲＭＱ，　特别注意存在 x > y的查询。

Accepted Code:

 1 /*************************************************************************
 2     > File Name: 3603.cpp
 3     > Author: Stomach_ache
 4     > Mail: sudaweitong@gmail.com
 5     > Created Time: 2014年07月10日 星期四 21时17分41秒
 6     > Propose: 
 7  ************************************************************************/
 8 
 9 #include <cmath>
10 #include <string>
11 #include <cstdio>
12 #include <fstream>
13 #include <cstring>
14 #include <iostream>
15 #include <algorithm>
16 using namespace std;
17 
18 const int MAX_N = 300002;
19 
20 int n, m;
21 int vis[MAX_N], len[MAX_N], rmq[MAX_N][25];
22 
23 void rmq_init() {
24       for (int i = 1; i <= n; i++) rmq[i][0] = len[i];
25     for (int j = 1; (1<<j) <= n; j++) {
26           for (int i = 1; i+(1<<(j-1))-1 <= n; i++) {
27               rmq[i][j] = max(rmq[i][j-1], rmq[i+(1<<(j-1))][j-1]);
28         }
29     }
30 }
31 
32 int RMQ(int L, int R) {
33       int k = (int)(log(R - L + 1.0) / log(2.0));     
34     return max(rmq[L][k], rmq[R-(1<<k)+1][k]);
35 }
36 
37 int solve(int x, int y) {
38       int  = x, R = y, ans = -1;
39     while (L <= R) {
40         int M = (L + R) / 2;
41         if (len[M] >= M - x + 1) {
42               L = M + 1;
43         } else {
44               ans = M;
45             R = M - 1;
46         }
47     }
48     if (ans == -1) return y - x + 1;
49     return max(ans - x, RMQ(ans, y));
50 }    
51 
52 int 
53 main(void) {
54     while (~scanf("%d", &n)) {
55           memset(vis, 0, sizeof(vis));
56           memset(len, 0, sizeof(len));
57         int begin = 1;
58           for (int i = 1; i <= n; i++) {
59               int tmp;
60             scanf("%d", &tmp);
61             len[i] = min(i-begin+1, i-vis[tmp]);
62             begin = max(begin, vis[tmp]+1);
63             vis[tmp] = i;
64         }
65         //for (int i = 1; i <= n; i++) printf("%d
", len[i]);
66         rmq_init();
67         scanf("%d", &m);
68         while (m--) {
69               int x, y;
70             scanf("%d %d", &x, &y);
71             // 测试数据可能会有 x > y　的情况
72             if (x > y) swap(x, y);
73             printf("%d
", solve(x, y));
74         }
75     
76     }
77 
78     return 0;
79 }