[hdu4763]next数组的应用

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：

 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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