POJ 3320 Jessica's Reading Problem

Jessica’s Reading Problem

Time Limit: 1000MS Memory Limit: 65536K

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

思路：

就是找，从1-n开始遍历，假如出现的知识点已经到全部涉及到了，那么就计算最小值，也就时记录的区间，最后的话 ，看看区间前面的点，是否在后面会出现，也就是重复的知识点，是的话就让前区间+1。

第一种做法：

#include <iostream>
#include <algorithm>
#include <set>
#include <map> 
using namespace std;
int a[1000010];
map<int, int> mp;
set<int> st;
int main() {
	int n;
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%d", &a[i]);
		st.insert(a[i]);
	}
	int ans = n + 1, sum = 0, t = 1;
	for (int i = 1; i <= n; i++) {
		if (mp[a[i]]++ == 0) sum++;
		if (sum == st.size()) {
			while (mp[a[t]] > 1) mp[a[t++]]--;
			ans = min(ans, i - t + 1);
		}
	}
	cout << ans << endl;
	return 0;
}

第二种做法：

思路差不多，实现有所差异

#include <iostream>
#include <algorithm>
#include <set>
#include <map> 
using namespace std;
int a[1000010];
map<int, int> mp;
set<int> st;
int main() {
	int n;
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%d", &a[i]);
		st.insert(a[i]);
	}
	int ans = n, sum = 0, t = 0, m = st.size(), s = 0;
	while (1) {
		while (t < n && sum < m) {
			if (mp[a[t++]]++ == 0) sum++;
		}
		if (sum < m) break;
		ans = min(ans, t - s);
		if (--mp[a[s++]] == 0) {
			sum--;
		}
	}
	cout << ans << endl;
	return 0;
}