【HDU 3336】Count the string 【KMP next数组巧妙应用】

最新推荐文章于 2022-10-26 15:00:00 发布

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最新推荐文章于 2022-10-26 15:00:00 发布

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分类专栏：总是不会的DP KMP

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本文介绍了一种利用KMP算法预处理和动态规划求解字符串中所有非空前缀出现次数之和的方法，并给出了一段C++代码实现。该问题需要处理大量字符串匹配情况，通过动态规划优化计算过程。

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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

居然可以用dp
dp[i]=dp[next[i]]+1 ;

#include<bits/stdc++.h>
using namespace std ;
typedef long long LL ;

const int MAXN = 200000+10 ;
const int MAXM = 1e5 ;
const int mod  =10007 ;

int m;
char str[MAXN];
int dp[MAXN],nexts[MAXN];
void getn(){
    int i,j;i=j=0;
    nexts[0]=-1;j=-1;
    while(i<m){
        if(j==-1||str[i]==str[j]) nexts[++i]=++j;
        else j=nexts[j];
    }
}

int main(){
    int t;scanf("%d",&t);
    while(t--){
        int n;scanf("%d",&n);
        scanf("%s",str);
        m=strlen(str);
        getn();
        int ans=0;
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=m;i++){
            dp[i]=dp[nexts[i]]+1;
            ans=(ans+dp[i]%mod)%mod;
        }
        printf("%d\n",ans);
    }

    return  0;
}