【HDU 3336】Count the string（KMP+DP）

最新推荐文章于 2025-08-06 16:23:50 发布

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最新推荐文章于 2025-08-06 16:23:50 发布

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文章标签： java 数据结构与算法

原文链接：http://www.cnblogs.com/zsyacm666666/p/7284018.html

本文探讨了一个经典的字符串匹配问题，通过KMP算法解决所有非空前缀的匹配次数计算，并提供了一种动态规划的方法来高效求解该问题。文章还包含了一个完整的Java实现示例。

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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

题解

考虑kmp算法中next数组的定义，即
\[f[i]=k\;whilea[0....k]==a[k+1...j]\]
那么我们就可以产生一个dp方程
设dp[i]：已a[i]结尾的前缀数
\[dp[i]=dp[f[i]]+1\]

参考代码

import java.io.*;
import java.util.*;

public class Main{
      static int N=200000+10;
      static int f[]=new int [N];
      static char a[]=new char[N];
      static void getFail(char b[],int m) {
          int j=0;
          f[1]=0;
          for(int i=2;i<=m;i++) {
              while(j>0&&b[j+1]!=b[i]) j=f[j];
              if(b[j+1]==b[i]) j++;
              f[i]=j;
          }
      }
      static int dp[]=new int[N];
      public static void main(String[] args){
          InputReader in=new InputReader(System.in);
          PrintWriter out=new PrintWriter(System.out);
          int T=in.nextInt();
          while(T--!=0) { 
             int n=in.nextInt();
             String str=in.next();
             for(int i=0;i<n;i++) a[i+1]=str.charAt(i);
             getFail(a,n);
             Arrays.fill(dp, 0);
             int ans=0;
             for(int i=1;i<=n;i++) {
                 dp[i]=(dp[f[i]]+1)%10007;
                 ans+=dp[i];
                 ans%=10007;
             }
             out.println(ans);
             out.flush();
          }
      }
    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;
        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }
        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }
        public int nextInt() {
            return Integer.parseInt(next());
        }
    }
}

转载于:https://www.cnblogs.com/zsyacm666666/p/7284018.html