Codeforces Round #116 (Div. 2, ACM-ICPC Rules) E. Cubes (尺取)

 1 //#pragma comment(linker, "/STACK:102400000, 102400000")
 2 #include <algorithm>
 3 #include <iostream>
 4 #include <cstdlib>
 5 #include <cstring>
 6 #include <cstdio>
 7 #include <vector>
 8 #include <cmath>
 9 #include <ctime>
10 #include <list>
11 #include <set>
12 #include <map>
13 using namespace std;
14 typedef long long LL;
15 typedef pair <int, int> P;
16 const int N = 1e5 + 5;
17 vector <int> ans[N];
18 int a[N*2];
19 
20 int solve(int u, int k) {
21     int len = ans[u].size(), notcnt = 0, cnt = 1, l = 0, res = 1;
22     //notcnt表示中间删除数的多少，cnt表示颜色为u的序列的长度
23     for(int i = 1; i < len; ++i) {
24         int v = ans[u][i];
25         notcnt += v - ans[u][i - 1] - 1; 
26         cnt++;
27         while(notcnt > k && l <= i) { 
28             notcnt -= ans[u][l + 1] - ans[u][l] - 1;
29             cnt--;
30             l++;
31         }
32         res = max(cnt, res);
33     }
34     return res;
35 }
36 
37 int main()
38 {
39     int n, m, k;
40     scanf("%d %d %d", &n, &m, &k);
41     for(int i = 1; i <= n; ++i) {
42         scanf("%d", a + i);
43         ans[a[i]].push_back(i);
44     }
45     int res = 1;
46     for(int i = 1; i <= m; ++i) {
47         if(ans[i].size()) { //要是存在颜色为i的数
48             res = max(res, solve(i, k));
49         }
50     }
51     printf("%d\n", res);
52     return 0;
53 }