| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 14734 | Accepted: 5073 |
Description
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 P non-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
Source
题意：一本书有P页，每一页都一个知识点，求去最少的连续页数覆盖所有的知识点。
分析：和上面的题一样的思路，如果一个区间的子区间满足条件，那么在区间推进到该处时，右端点会固定，左端点会向右移动到其子区间，且其子区间会是更短的，只是需要存储所选取的区间的知识点的数量，那么使用map进行映射以快速判断是否所选取的页数是否覆盖了所有的知识点。
可以用map来查重，计数。
居然可以写那么简洁
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <map> 5 #define inf 0x3f3f3f3f 6 using namespace std; 7 int a[1000005]; 8 int main() 9 { 10 map<int, int>m; 11 int n; 12 scanf("%d", &n); 13 int i; 14 int sum; 15 sum = 0; 16 for (i = 1; i <= n; i++) 17 { 18 scanf("%d", &a[i]); 19 if (!m[a[i]]) 20 { 21 sum++; 22 m[a[i]]++; 23 } 24 } 25 m.clear(); 26 int st = 1; 27 int ed = 1; 28 int s = 0; 29 int ans = inf; 30 while (1) 31 { 32 while (ed <= n && s < sum)//是否超过边界或者未达到 33 { 34 if (!m[a[ed++]]++) s++;//ed向前，如果曾经没有或为0，s++ 35 } 36 if (s < sum) break; 37 ans = min(ans, ed - st); 38 if (!(--m[a[st++]])) s--;//st向前，如果曾经没有或为0，s-- 39 } 40 printf("%d ", ans); 41 return 0; 42 }