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 }

转载于:https://www.cnblogs.com/Running-Time/p/4550274.html

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