hdu 3336（KMP算法的应用）

最新推荐文章于 2020-11-18 21:41:36 发布

柠檬与茉莉

最新推荐文章于 2020-11-18 21:41:36 发布

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分类专栏： KMP 文章标签： acm kmp

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本文介绍了一道经典的字符串匹配问题——HDU 3336 Count the String，并提供了详细的解题思路及C++代码实现。通过分析字符串的前缀匹配次数来计算总匹配数，利用next数组进行高效求解。

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3336

Count the string

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4844 Accepted Submission(s): 2294

Problem Description

It is well known that AekdyCoin is good at string problems as well as number theory problems. When given a string s, we can write down all the non-empty prefixes of this string. For example:
s: "abab"
The prefixes are: "a", "ab", "aba", "abab"
For each prefix, we can count the times it matches in s. So we can see that prefix "a" matches twice, "ab" matches twice too, "aba" matches once, and "abab" matches once. Now you are asked to calculate the sum of the match times for all the prefixes. For "abab", it is 2 + 2 + 1 + 1 = 6.
The answer may be very large, so output the answer mod 10007.

Input

The first line is a single integer T, indicating the number of test cases.
For each case, the first line is an integer n (1 <= n <= 200000), which is the length of string s. A line follows giving the string s. The characters in the strings are all lower-case letters.

Output

For each case, output only one number: the sum of the match times for all the prefixes of s mod 10007.

Sample Input

1
4
abab

Sample Output

思路：要想解出这道题就得先理解next[]的求法和含义；

#include <iostream>
#include <string.h>
#include <string>
#include <stdio.h>
const int N=200000+100;
const int mod=10007;
using namespace std;
int n,next[N];
char s[N];

void get_next() //求出next[]
{
  int i=1,j=0;
  while(i<=n)
  {
    if(j==0||s[i]==s[j])
    {
      ++i;
      ++j;
      next[i]=j;
    }
    else
      j=next[j];
  }
}

int main()
{
     int T;
     cin>>T;
     while(T--)
     {
      scanf("%d",&n);

      scanf("%s",s+1);
      get_next();
      int cnt=0;
      cnt=(cnt+n)%mod;

      for(int i=3;i<=n+1;i++)
      {
        if(next[i]>=2)
        {
          cnt=(cnt+1)%mod;
        }
      }
      printf("%d\n",cnt);
     }
     return 0;
}