尺取法 POJ 3320 Jessica's Reading Problem

题目传送门

 /*
     尺取法：先求出不同知识点的总个数tot，然后以获得知识点的个数作为界限， 更新最小值
 */
 #include <cstdio>
 #include <cmath>
 #include <cstring>
 #include <algorithm>
 #include <set>
 #include <map>
 using namespace std;

 const int MAXN = 1e6 + ;
 const int INF = 0x3f3f3f3f;
 int a[MAXN];

 int main(void)        //POJ 3320 Jessica's Reading Problem
 {
     int n;
     while (scanf ("%d", &n) == )
     {
         set<int> S;
         for (int i=; i<=n; ++i)
         {
             scanf ("%d", &a[i]);    S.insert (a[i]);
         }

         map<int, int> cnt;
         int tot = S.size ();    int ans = n, num = ;    int i = , j = ;
         while ()
         {
             while (j <= n && num < tot)    if (cnt[a[j++]]++ == )    num++;
             if (num < tot)    break;
             ans = min (ans, j - i);
             if (--cnt[a[i++]] == ) num--;
         }

         printf ("%d\n", ans);
     }

     return ;
 }