HDU 3336 Count the string(KMP)||记忆化搜索_kmp记忆化搜索-优快云博客

本文介绍了一种计算字符串所有非空前缀在原字符串中匹配次数的方法，并提供了两种不同的实现方案，一种是使用KMP算法的改进版本进行动态规划，另一种是通过记忆化搜索优化匹配过程。

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Count the string

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4105 Accepted Submission(s): 1904

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

6

目前对KMP算法还是不熟练，硬套模板，缺乏思考。。

先求next值，然后DP。

　sum[]数组：以 i 结尾的串中所有前缀的计数和
状态转移方程：sum[i]=(sum[next[i]]+1)%mod

#include <cstdio>
#include <iostream>
#include<cstring>
using namespace std;
#define maxn 200010
#define mod 10007
char a[maxn];
int next[maxn],len,sum[maxn];
void get_next()
{
    int count = 0 ;
    int j = -1 ;
    int i = 0;
    next[0]=-1;
    while( i < len)
    {
        if(-1 == j || a[i] == a[j])
        {
            i++;
            j++;
//            if(a[i] != a[j])
//            {
                next[i] = j;
//            }
//            else {
//                next[i]= next[j];
//            }
        }
        else
            j = next[j];
    }
}
int main()
{
    int n,i;
    scanf("%d",&n);
    while(n--)
    {
        int ans = 0;
        scanf("%d",&len);
        scanf("%s",a);
        get_next();
        sum[0]=0;
        for(i = 1; i <= len ;i++)
        {
            sum[i]=(sum[next[i]]+1)%mod;
            ans = (ans + sum[i])%mod;
        }
        printf("%d\n",ans);
    }
    return 0;
}

可以采用记忆化搜索，用数组b来记录与前一个前缀串匹配的所有字串的最后一个字符。

第一次时记录下b，以后每次只要用a[b[j]+1]与最后一个字符去匹配。

#include<stdio.h>
#define maxn 200006
#define mod 10007
char a[maxn];
int b[maxn];
int main()
{
    int t,len,j,i;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&len);
        scanf("%s",a);
        char c  = a[0];
        int k = 0 ,count = 0 ;
        for(i = 0; i < len ; i++){
            if(a[i]==c)
            {
                count ++ ;
                count%=mod;
                b[k++]=i;
            }
        }
//首先记录第一个字符出现的位置,并保存次数
//        for(i = 0 ; i < k ; i++)
//        printf("%d ",b[i]);
//        puts("");
        for(i = 1 ; i < len ;i++)
        {
            int l= 0 ;
            for(j = 0 ; j < k ; j++)
            {
                if(a[i] == a[b[j]+1])//直接比较已经匹配的上一个位置，是否匹配下一个字符
                {
                    count ++;
                    count %= mod;
                    b[l++]=b[j]+1;
                }
            }
            k = l;
        }
        printf("%d\n",count);
    }
    return 0;
}