Count the string(记忆化搜索)

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#记忆化搜索

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本文介绍了一个关于字符串的算法问题，任务是计算一个字符串中所有非空前缀出现的次数总和，并通过模运算处理可能的大数值结果。文章提供了一种使用记忆化搜索的方法来解决该问题，并附带了完整的C++实现代码。

Count the string

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4899 Accepted Submission(s): 2313

题目连接

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

Author

foreverlin@HNU

Source

HDOJ Monthly Contest – 2010.03.06

记忆化搜索: 预先记录前一状态的位置.

AC代码:

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;

const int M = 1e5 * 2 + 100;
char s[M];
int map[M];

int main()
{
    int T,n;
    scanf("%d",&T);
    while(T--)
    {
        memset(map,0,sizeof(map));
        scanf("%d %s",&n,s);
        int tp,len,ans;
        len = strlen(s);
        map[0] = ans = 0;
        tp = 1;
        for(int i = 1; s[i]; i++)
            if(s[i] == s[0]) map[tp++] = i;
        ans += tp;
        for(map[0]++; map[0] < len; map[0]++)
        {
            int temp = 1;
            for(int i = 1; i < tp; i++)
                if(s[map[i] + 1] == s[map[0]]) map[temp++] = map[i] + 1;
            tp = temp;
            ans += tp;
            if(ans > 10007) ans %= 10007;
        }
        printf("%d\n",ans);
    }
    return 0;
}