技術標籤：學習筆記演算法c++

scanf比cin的效率高出一倍還要多

上面A的兩次除了資料讀取不一樣其餘完全相同！！！

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

題意：一本書總共有P頁，第i頁恰好有知識點ai，全書中一個知識點可能會多次提到
找一個頁碼範圍，把所有知識點都涵蓋。輸出最小頁數
和前面那個Subsequence一樣，尺取法。
所有知識點都覆蓋既每個知識點出現的次數都不小於1

#include<iostream>
#include<string>
#include<set>
#include<map>
using namespace std;
#define MAX 1000000
int a[MAX+5];
set<int> all;
map<int, int> mycount;
//std中有count
int main()
{
	int p; cin >> p;
	//用set計算知識點的個數n
	
	int n;
	for (int i = 0; i < p; i++) {
		//cin >> a[i];
		scanf("%d", &a[i]);
		all.insert(a[i]);
	}
	n = all.size();

	int left = 0, right = 0, num = 0;//num 記錄這個區間包含的知識點數
	int result = p+1;
	//鍵是知識點，值是該知識點出現的次數
	for(;;)
	{
		while(right<p&&num<n)
		{
			if (mycount[a[right++]]++ == 0) {
				//count[a[right]]++; 知識點a[right] 只要出現 其count就應該++
				num++;
			}
		}
		if (num < n)break;
		result = min(result, right - left);
		//移動左端點
		if (--mycount[a[left++]] == 0)num--;//一個知識點沒了
	}
	cout << result << endl;
	return 0;
}