尺取法 POJ 3320 Jessica's Reading Problem

 

题目传送门

 1 /*  2  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  3 */  4 #include <cstdio>  5 #include <cmath>  6 #include <cstring>  7 #include <algorithm>  8 #include <set>  9 #include <map> 10 using namespace std; 11 12 const int MAXN = 1e6 + 10; 13 const int INF = 0x3f3f3f3f; 14 int a[MAXN]; 15 16 int main(void) //POJ 3320 Jessica's Reading Problem 17 { 18 int n; 19 while (scanf ("%d", &n) == 1) 20  { 21 set<int> S; 22 for (int i=1; i<=n; ++i) 23  { 24 scanf ("%d", &a[i]); S.insert (a[i]); 25  } 26 27 map<int, int> cnt; 28 int tot = S.size (); int ans = n, num = 0; int i = 1, j = 1; 29 while (1) 30  { 31 while (j <= n && num < tot) if (cnt[a[j++]]++ == 0) num++; 32 if (num < tot) break; 33 ans = min (ans, j - i); 34 if (--cnt[a[i++]] == 0) num--; 35  } 36 37 printf ("%d\n", ans); 38  } 39 40 return 0; 41 }