[hdu4763]next数组的应用

时间:2023-01-03 17:44:06

http://acm.hdu.edu.cn/showproblem.php?pid=4763

题目大意:给一个字符串,判断是否可以写成ABACA,B、C表示长度大于等于0的字符串。

方法:ans = next[len]如果小于等于len/3,则ans是最大可能的答案,否则ans = next[ans] 要继续往前走,直到ans <= len/3, 然后枚举从2*ans位置枚举到len - ans,看是否存在某个位置j, 使得next[j] = ans,当然这里也有一个往前走的过程(如果next[j]一开始大于ans),具体见代码:

[hdu4763]next数组的应用[hdu4763]next数组的应用
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <map>
 7 #include <vector>
 8 #include <stack>
 9 #include <string>
10 #include <ctime>
11 #include <queue>
12 #define mem0(a) memset(a, 0, sizeof(a))
13 #define mem(a, b) memset(a, b, sizeof(a))
14 #define lson l, m, rt << 1
15 #define rson m + 1, r, rt << 1 | 1
16 #define eps 0.0000001
17 #define lowbit(x) ((x) & -(x))
18 #define memc(a, b) memcpy(a, b, sizeof(b))
19 #define x_x(a) ((a) * (a))
20 #define LL long long
21 #define DB double
22 #define pi 3.14159265359
23 #define MD 10000007
24 #define INF (int)1e9
25 #define max(a, b) ((a) > (b)? (a) : (b))
26 using namespace std;
27 char str[1200000];
28 int next[1200000];
29 void getNext()
30 {
31         next[0] = next[1] = 0;
32         for(int i = 1; str[i]; i++) {
33                 int j = next[i];
34                 while(j && str[i] != str[j]) j = next[j];
35                 next[i + 1] = str[i] == str[j]? j + 1 : 0;
36         }
37 }
38 int solve()
39 {
40         int len = strlen(str), ans = next[len], F = 0;
41         while(ans * 3 > len) ans = next[ans];
42         while(ans) {
43                 for(int i = ans * 2; i <= len - ans; i++) {
44                         int j = next[i];
45                         while(j > ans) j = next[j];
46                         if(j == ans) {
47                                 F = 1;
48                                 break;
49                         }
50                 }
51                 if(F) break;
52                 ans = next[ans];
53         }
54         return ans;
55 }
56 int main()
57 {
58         //freopen("input.txt", "r", stdin);
59         int T;
60         cin>> T;
61         for(int i = 1; i <= T; i++) {
62                 scanf("%s", str);
63                 getNext();
64                 cout<< solve()<< endl;
65         }
66         return 0;
67 }
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