POJ3320尺取法

Jessica's Reading Problem

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 9249 Accepted: 3001

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

51 8 8 8 1

Sample Output

2

题目大意：

       看最小的页数，获得整本书的所有知识点。求这个页数是多少。

遇到的问题和思路：

       挑战程序竞赛149的例题。个人在这里先整理一下思路：①用利用set集合里面元素的唯一性，将a[i]的所有元素放到里面去，这样就知道了一共有几个知识点。②尺取法的原理是从起始left到末端right中寻找答案，然后利用map的映射，获得每个知识点所出现的数目即可。③退出条件是sun < 知识点数目。

给出代码：

#include<cstdio>#include#include<cstring>#include<set>#include<map>using namespace std;int p;int a[1000000 + 10];void solve(){set <int> ind;for (int i = 0; i < p; i++){ind.insert(a[i]);}int left = 0, right = 0, sum = 0;map <int int=""> ans;int res = p;//假设全部都看一遍while (true){while (right < p && sum < ind.size()){//printf("ans[] = %d\n", ans[a[right]]);if (ans[a[right++]]++ == 0){sum++;}}//printf("sum = %d\n", sum);if (sum < ind.size()) break;res = min(res, right - left);if (--ans[a[left++]] == 0){//注意，这里不论if条件是否成立，--ans[a[left++]]的加减都会执行sum--;}} printf("%d\n", res);}int main(){while(scanf("%d", &p) != EOF){memset(a, 0, sizeof(a));for (int i = 0; i < p; i++){scanf("%d", a + i);}solve();}return 0;}</int></int></map></set></cstring></algorithm></cstdio>