POJ - 3320 Jessica's Reading Problem

Jessica's a very lovely girl wooed by lots of boys. Recently she has a problem. The final exam is coming, yet she has spent little time on it. If she wants to pass it, she has to master all ideas included in a very thick text book. The author of that text book, like other authors, is extremely fussy about the ideas, thus some ideas are covered more than once. Jessica think if she managed to read each idea at least once, she can pass the exam. She decides to read only one contiguous part of the book which contains all ideas covered by the entire book. And of course, the sub-book should be as thin as possible.

A very hard-working boy had manually indexed for her each page of Jessica's text-book with what idea each page is about and thus made a big progress for his courtship. Here you come in to save your skin: given the index, help Jessica decide which contiguous part she should read. For convenience, each idea has been coded with an ID, which is a non-negative integer.

Input

The first line of input is an integer P (1 ≤ P≤ 1000000), which is the number of pages of Jessica's text-book. The second line contains Pnon-negative integers describing what idea each page is about. The first integer is what the first page is about, the second integer is what the second page is about, and so on. You may assume all integers that appear can fit well in the signed 32-bit integer type.

Output

Output one line: the number of pages of the shortest contiguous part of the book which contains all ideals covered in the book.

Sample Input

5
1 8 8 8 1

Sample Output

2

 1 /*
 2 很显然一共有5页书，却只有2个知识点，想读就读前两页，所以就输出2页
 3 先想想需要解决什么问题：
 4 1.要解决出现一共多少个知识点
 5 2.根据当前子序列包含的知识点数来入队、出队,其实就这样
 6 第一个问题其实可以用set来解决，知识点数量就是set的size大小。
 7 下一个问题其实可以用map来存，存每一页出现的知识点在这个子序列中出现的次数
 8 如果是0的话就把这个知识点放进来并++，直到所有知识点覆盖为止。
 9 在所有知识点覆盖的基础上出队，直到某一个知识点在子序列的出现次数为0的时候
10 再从后面入队，并在这些操作中记录下最少页数就可以了，剩下的和第一颗栗子一个味道
11 
12 */
13 #include <stdio.h>
14 #include <set>
15 #include <map>
16 #include <iostream>
17 #include <algorithm>
18 using namespace std;
19 int P, n, a[100000 + 10];
20 void solved() {
21     set<int> all; //set容器不允许相同的元素出现
22     for (int i = 0; i < P; i++) {
23         all.insert(a[i]);
24     }
25     int n = all.size();
26     int s = 0, t = 0, num = 0;
27     map<int, int> count;// key-知识点，value-出现次数
28     int res = P;
29     while (true) {
30         while (t < P && num < n) {
31             //出现了一个新的知识点
32             if (count[a[t++]]++ == 0)
33                 num++;
34         }
35         if (num < n) break;
36         res = min(res, t - s);
37         if (--count[a[s++]] == 0)
38             num--;
39     }
40     printf("%d\n", res);
41 }
42 int main() {
43     while (scanf("%d", &P) != EOF) {
44         for (int i = 0; i < P; i++) {
45             scanf("%d", &a[i]);
46         }
47         solved();
48     }
49     return 0;
50 }

 

转载于:https://www.cnblogs.com/z-712/p/7324545.html