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UVA 11572 Unique Snowflakes (贪心技巧)

热度:15   发布时间:2023-11-15 16:46:37.0

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

#include<algorithm>
#include<set>
#include<cstdio>
#define maxn 1000005
using namespace std;int n,A[maxn];
/*
题目大意:给定n个数,
找出不含有相同两个数的最长连续段的长度。本题思维很巧妙,当然我也是抄紫书上的代码,
跟最长连续子序列的道理一样,
需要维护一个可选答案,然后不断扩充,
直到所有最大的可能答案都筛选过了。
这时需要set集合维护不可重集,
和count函数。
*/
int main()
{int t;scanf("%d",&t);while(t--){scanf("%d",&n);for(int i=0 ; i<n ; i++)  scanf("%d",&A[i]);set<int> s;int L=0,R=0,ans=0;while(R<n){while(R<n && !s.count(A[R])) s.insert(A[R++]);ans=max(ans,R-L);s.erase(A[L++]);}//op=max(e-s+1,op);printf("%d\n",ans);}return 0;
}

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