HDU 2756 & UVA 11572 Unique Snowflakes

Problem Description

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. 

 

 

Input

The first line of each test chunk contains an integer specifying the number of test cases in this chunk 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. 
Please process to the end of the data file.

 

 

Output

For each test case output a line containing single integer, the maximum number of unique snowflakes that can be in a package. 

 

思路类似XDOJ1175，用dp[i]记录以i结尾的最长不重复序列，用m[a[i]]记录上一次a[i]出现的下标

PS:HDU的输入与原题略有不同

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <map>
#include <bitset>
using namespace std;
#define SIZE 200005
typedef long long LL;
int dp[SIZE],a[SIZE],n;
map<LL,int> m;
int main(){
  // freopen("test.in","r",stdin);
  ios::sync_with_stdio(false);
  int T;
  while (cin >> T ) {
  for (int times = 1; times <= T; times ++){
    m.clear();
    memset(dp,0,sizeof(dp));
    cin >> n;
    for (int i=1;i<=n;i++){
      cin >> a[i];
    }
    dp[1] = 1; m[a[1]] = 1;
    for (int i=2;i<=n;i++){
      if (m[a[i]] == 0){
        dp[i] = dp[i-1] + 1; m[a[i]] = i;
      }
      else {
        dp[i] = i - m[a[i]];
        m.clear();
        for (int j=i,k=0;k<dp[i];k++){
          m[a[j-k]] = j - k;
        }
      }
    }
    int res = 0;
    for (int i=1;i<=n;i++){
      res = max(res,dp[i]);
      // cout << dp[i] << endl;
    }
    cout << res << endl;
  }
  }
  return 0;
}