HDU 3336 Count the string（kmp + dp）

妖狐艹你老母 2022-09-30 03:50 148阅读 0赞

Count the string

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)  
Total Submission(s): 9715 Accepted Submission(s): 4543

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

Author  
foreverlin@HNU

Source  
HDOJ Monthly Contest – 2010.03.06  
传送门：[点这里][Link 1]

假设两串字符完全相等，next\[j\]=i,代表s\[1…i\]==sum\[j-i+1….j\]，这一段其实就是前缀

i~j之间已经不可能有以j结尾的子串是前缀了，不然next【j】就不是 i 了

设dp【i】:以string\[i\]结尾的子串总共含前缀的数量

所以dp\[j\]=dp\[i\]+1，即以i结尾的子串中含前缀的数量加上前j个字符这一前缀

#include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    
    using namespace std;
    const int maxn = 2e5 + 10;
    const int MOD = 10007;
    
    void get_next(char *t,int *next)//next数组模板
    {
       int i=0;//Prefix 前缀
       int j=1;//Postfix 后缀
       next[0] = 0;//自定义的，0和1都从0开始匹配
       next[1] = 0;
       while(t[j] != '\0')
       {
           if(t[i] == t[j])//若前后字符匹配，则向前推进
           {
               i++;
               j++;
               next[j] = i;
    
           }
           else
           {
               i = next[i];///前后字符不匹配，则回溯。注意，此时是i和j不匹配，因此，根据next数组定义，要回溯到next[i]的值
           }
           if(0 == i)//当回溯到首字符时，单独进行处理
           {
               if(t[i] == t[j])
               {
                   next[++j] = ++i;
               }
               else next[++j] = i;
           }
       }
    }
    
    int main()
    {
        int T;
        cin>>T;
        while(T--)
        {
            int n;
            cin>>n;
            char s[maxn];
            cin>>s;
            int next[maxn];
            get_next(s,next);
            int d[maxn];
            memset(d,1,sizeof(d));
            d[0]=0;
            int sum=0;
            for(int i=1;i<=n;i++)
            {
                d[i]=d[next[i]]+1;
                sum += d[i]%10007;
            }
            cout<<sum%10007<<endl;
        }
        return 0;
    }

[Link 1]: http://acm.hdu.edu.cn/showproblem.php?pid=3336