Emily the entrepreneur has a cool business idea: packaging and selling snow?akes. She has devised a machine that captures snow?akes as they fall, and serializes them into a stream of snow?akes that ?ow, 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 snow?ake in a package must be di?erent from the others. Unfortunately, this is easier said than done, because in reality, many of the snow?akes ?owing through the machine are identical. Emily would like to know the size of the largest possible package of unique snow?akes that can be created. The machine can start ?lling the package at any time, but once it starts, all snow?akes ?owing 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 snow?akes have ?owed out of the machine.
Input
The ?rst 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 snow?akes processed by the machine. The following n lines each contain an integer (in the range 0 to 109, inclusive) uniquely identifying a snow?ake. Two snow?akes are identi?ed by the same integer if and only if they are identical. The input will contain no more than one million total snow?akes.
Output
For each test case output a line containing single integer, the maximum number of unique snow?akes that can be in a package.
Sample Input
1 5 1 2 3 2 1
Sample Output
3
分析：
从一个有n个数的序列中，找到一个最大的子段（连续的），使得这个最大子段中没有重复元素，输出最大子段的长度。
注意：序列是不连续的，子段必须是连续的
做法：
对于该类段查找问题可以采用经典的滑动窗口方法，即维护一个窗口，窗口的左右边界用两个变量L,R代表，先增加R直到出现重复数字，再增加L，再增加R，直到R达到n
贴张图形象得一批~
set好用！
code：
#include<stdio.h>
#include <iostream>
#include <math.h>
#include <queue>
#include<set>
using namespace std;
#define max_v 1000005
int a[max_v];
int main()
{
set<int> s;
int t,n;
cin>>t;
while(t--)
{
cin>>n;
for(int i=0;i<n;i++)
cin>>a[i];
s.clear();
int left=0,right=0,ans=0;
while(right<n)
{
while(right<n&&!s.count(a[right]))
{
s.insert(a[right++]);
}
ans=max(ans,right-left);
s.erase(a[left++]);
}
cout<<ans<<endl;
}
return 0;
}
原文地址：https://www.cnblogs.com/yinbiao/p/9397004.html