UVA - 11536 - Smallest Sub-Array（滑动窗口）

题意：从给定字符串中，选最短连续子序列，包含1~k中的所有数。

尺取法（滑动窗口）解决：在第一次找到1~k的序列之后，向右滑动保证每个数至少存在一次，不断取最小值即可。

 

 1 #include<cstdio>  
 2 #include<cstring>  
 3 #include<cctype>  
 4 #include<cstdlib>  
 5 #include<cmath>  
 6 #include<iostream>  
 7 #include<sstream>  
 8 #include<iterator>  
 9 #include<algorithm>  
10 #include<string>  
11 #include<vector>  
12 #include<set>  
13 #include<map>  
14 #include<deque>  
15 #include<queue>  
16 #include<stack>  
17 #include<list>  
18 typedef long long ll;  
19 typedef unsigned long long llu;  
20 const int MAXN = 100 + 10;  
21 const int MAXT = 1000000 + 10;  
22 const int INF = 0x7f7f7f7f;  
23 const double pi = acos(-1.0);  
24 const double EPS = 1e-6;  
25 using namespace std;  
26   
27 int n, m, k, a[MAXT], T, vis[MAXN];  
28   
29 void init(){  
30     a[0] = 1, a[1] = 2, a[2] = 3;  
31     for(int i = 3; i < n; ++i)  a[i] = (a[i - 1] + a[i - 2] + a[i - 3]) % m + 1;  
32 }  
33   
34 int solve(){  
35     int head = 0, num = 0, ans = INF;  
36     memset(vis, 0, sizeof vis);  
37     for(int tail = 0; tail < n; ++tail)  
38         if(a[tail] >= 1 && a[tail] <= k){  
39             if(!vis[a[tail]])  ++num;  
40             ++vis[a[tail]];  
41             if(num == k){  
42                 while((a[head] >= 1 && a[head] <= k && vis[a[head]] > 1) || a[head] < 1 || a[head] > k){  
43                     if(a[head] >= 1 && a[head] <= k && vis[a[head]] > 1)  --vis[a[head]];  
44                     ++head;  
45                 }  
46                 ans = min(tail - head + 1, ans);  
47             }  
48         }  
49     return ans == INF ? -1 : ans;  
50 }  
51   
52 int main(){  
53     int ca = 0;  
54     scanf("%d", &T);  
55     while(T--){  
56         scanf("%d%d%d", &n, &m, &k);  
57         init();  
58         int ans = solve();  
59         if(ans != -1)  printf("Case %d: %d\n", ++ca, ans);  
60         else  printf("Case %d: sequence nai\n", ++ca);  
61     }  
62     return 0;  
63 }

 

posted @ 2016-10-25 10:56  TianTengtt  阅读(128)  评论(0)    收藏  举报