POJ3320 Jessica's Reading Problem 尺取法

Jessica’s Reading Problem、

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 22716 Accepted: 7687

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

題目意思就是要讀最少的連續書頁，複習到所有的知識點。
這道題目跟 poj3061 有點像，但是比之稍複雜。

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<map>
using namespace std;
const int N = 1e6 + 10;
map<int, int> m, Count;//第一個m是記錄書的總頁數。第二個是用來後面計數的
int a[N], n, ans, sum;
int solve() {
	int now = 0,l = 1, r = 1;
	while(true) {
		while(now < sum && r <= n) {//條件滿足或者後面沒有元素了
			if(Count[a[r++]]++ == 0)//這是第一次加入這本書。
				now++;
		}
			if(now < sum)
				break;
			ans = min(ans, r - l);//取答案的最下值。
			if(--Count[a[l++]] == 0)//這是裏面的最後一本書。
				now--;
	}
	return ans;
}
int main() {
	while(scanf("%d", &n) != EOF) {
		m.clear();//隊組讀入注意清零。
		Count.clear();
		ans = N, sum = 0;
		for(int i = 1; i <= n; i++) {
			scanf("%d", &a[i]);
			if(m[a[i]]++ == 0)
				sum++;
		}
		printf("%d\n", solve());
	}
	return 0;
}