uva 11572 Unique Snowflakes

原題：
Emily the entrepreneur has a cool business idea: packaging and selling snowflakes. She has devised a
machine that captures snowflakes as they fall, and serializes them into a stream of snowflakes that flow,
one by one, into a package. Once the package is full, it is closed and shipped to be sold.
The marketing motto for the company is “bags of uniqueness.” To live up to the motto, every
snowflake in a package must be different from the others. Unfortunately, this is easier said than done,
because in reality, many of the snowflakes flowing through the machine are identical. Emily would like
to know the size of the largest possible package of unique snowflakes that can be created. The machine
can start filling the package at any time, but once it starts, all snowflakes flowing from the machine
must go into the package until the package is completed and sealed. The package can be completed
and sealed before all of the snowflakes have flowed out of the machine.
Input
The first line of input contains one integer specifying the number of test cases to follow. Each test
case begins with a line containing an integer n, the number of snowflakes processed by the machine.
The following n lines each contain an integer (in the range 0 to 10 9 , inclusive) uniquely identifying a
snowflake. Two snowflakes are identified by the same integer if and only if they are identical.
The input will contain no more than one million total snowflakes.
Output
For each test case output a line containing single integer, the maximum number of unique snowflakes
that can be in a package.
Sample Input
1
5
1
2
3
2
1
Sample Output
3
題目大意：
給你n個數，問你最長的連續的不出現重複的數的長度是多少。

#include <bits/stdc++.h>using namespace std;int a[1000001];fstream in,out;map<int,int> mi;int main(){    ios::sync_with_stdio(false);    int t,n,ans;//  in.open("in.txt");//  out.open("answer.txt");    cin>>t;    while(t--)    {        cin>>n;        ans=1;        mi.clear();        for(int i=1;i<=n;i++)            cin>>a[i];        int l=1;        mi[a[1]]=1;        for(int i=2;i<=n;i++)        {            int mark=mi[a[i]];            if(mark>=l)                l=mark+1;            ans=max(ans,i-l+1);            mi[a[i]]=i;        }        cout<<ans<<endl;    }//  in.close();//  out.close();    return 0;}

解答:
在書上隨便翻到的一個例題，在leetcode裡面做過類似的。用一個map記錄每個數字出現過的位置，設定l和r分別表示這個最長連續不重複數組的兩端，其中r可以用遍曆中的i代替。如果遍曆到map尋找遍曆到的a[i]在前面出現的位置，如果在正在遍曆的數組當中就把l設定成上一次出現a[i]的位置並加一，不斷更新最大值即可。