xtu字符串 C. Marlon's String

最新推荐文章于 2023-08-15 12:24:37 发布

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最新推荐文章于 2023-08-15 12:24:37 发布

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文章标签： java

原文链接：http://www.cnblogs.com/crackpotisback/p/3877534.html

本文介绍了一个字符串匹配问题，即寻找所有可能的四元组(a,b,c,d)，使得两个字符串S的子串Sa..b与Sc..d连接后等于第三个字符串T。通过使用KMP算法进行前后缀匹配，并给出了一段实现该算法的C++代码。

C. Marlon's String

Time Limit: 2000ms

Memory Limit: 65536KB

64-bit integer IO format: %lld Java class name: Main

Long long ago, there was a coder named Marlon. One day he picked two string on the street. A problem suddenly crash his brain...

Let Si..j denote the i-th character to the j-th character of string S.

Given two strings S and T. Return the amount of tetrad (a,b,c,d) which satisfy Sa..b + Sc..d = T , a≤b and c≤d.

The operator + means concate the two strings into one.

Input

The first line of the data is an integer Tc. Following Tc test cases, each contains two line. The first line is S. The second line is T. The length of S and T are both in range [1,100000]. There are only letters in string S and T.

Output

For each test cases, output a line for the result.

Sample Input

1
aaabbb
ab

Sample Output

9

解题：与扩展KMP无关，与前缀有关。。。直接KMP

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <climits>
 7 #include <vector>
 8 #include <queue>
 9 #include <cstdlib>
10 #include <string>
11 #include <set>
12 #define LL long long
13 #define INF 0x3f3f3f3f
14 using namespace std;
15 int fail[100010];
16 char pa[100010],s[100010];
17 int a[100010],b[100010];
18 void kmp(int *arr){
19     int i,j,k;
20     fail[0] = fail[1] = 0;
21     for(i = 1; pa[i]; i++){
22         j = fail[i];
23         while(j && pa[i] != pa[j]) j = fail[j];
24         fail[i+1] = pa[j] == pa[i] ? j+1:0;
25     }
26     for(j = i = 0; s[i]; i++){
27         while(j && s[i] != pa[j]) j = fail[j];
28         if(pa[j] == s[i]){
29             arr[++j]++;
30         }
31     }
32     for(i = strlen(pa); i >= 0; i--)
33         if(fail[i]) arr[fail[i]] += arr[i];
34 }
35 int main(){
36     int t,i,len;
37     LL ans;
38     scanf("%d",&t);
39     while(t--){
40         scanf("%s %s",s,pa);
41         memset(a,0,sizeof(a));
42         memset(b,0,sizeof(b));
43         kmp(a);
44         reverse(s,s+strlen(s));
45         reverse(pa,pa+strlen(pa));
46         kmp(b);
47         ans = 0;
48         for(len = strlen(pa),i = 0; i < len; i++)
49             ans += (LL)a[i]*b[len-i];
50         printf("%lld\n",ans);
51     }
52     return 0;
53 }