HDU 6153 A Secret（擴展KMP算法）

版權聲明：可以轉載，但需註明出處，謝謝。 https://blog.csdn.net/Zhao_Xinhao/article/details/77450593

原題

A Secret

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 256000/256000 K (Java/Others)

Problem Description

Today is the birthday of SF,so VS gives two strings S1,S2 to SF as a present,which have a big secret.SF is interested in this secret and ask VS how to get it.There are the things that VS tell:
Suffix(S2,i) = S2[i...len].Ni is the times that Suffix(S2,i) occurs in S1 and Li is the length of Suffix(S2,i).Then the secret is the sum of the product of Ni and Li.
Now SF wants you to help him find the secret.The answer may be very large, so the answer should mod 1000000007.

Input

Input contains multiple cases.
The first line contains an integer T,the number of cases.Then following T cases.
Each test case contains two lines.The first line contains a string S1.The second line contains a string S2.
1<=T<=10.1<=|S1|,|S2|<=1e6.S1 and S2 only consist of lowercase ,uppercase letter.

Output

For each test case,output a single line containing a integer,the answer of test case.
The answer may be very large, so the answer should mod 1e9+7.

Sample Input

2
aaaaa
aa
abababab
aba

Sample Output

13
19

Hint

case 2:
Suffix(S2,1) = "aba",
Suffix(S2,2) = "ba",
Suffix(S2,3) = "a".
N1 = 3,
N2 = 3,
N3 = 4.
L1 = 3,
L2 = 2,
L3 = 1.
ans = （3*3+3*2+4*1）%1000000007.

Source

2017中國大學生程序設計競賽 - 網絡選拔賽

題意

在母串中找模式串的所有後綴的匹配次數，並乘上各種後綴對應長度，再把所有的結果加起來輸出。

涉及知識及算法

吐槽：開始以爲是KMP，做了很久都做不對。最後發現KMP會跳過一些情況，用擴展KMP是可以做的，但用KMP也不是不行，大神代碼傳送門：點擊打開鏈接

擴展KMP算法 傳送門：點擊打開鏈接

把兩個字符串都反轉一下就變成了前綴匹配，就可以利用擴展KMP的性質了。因爲extend[i] 表示的是從i到結尾的主串與模式串的最長公共前綴，這樣遍歷一遍主串，每一位對答案的貢獻就是1+⋯+extend[i]=extend[i]×(extend[i]+1)/2，累加即可。

代碼

#include <iostream>
#include <string>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MOD = 1000000007;
const int MAXL=1e6+5;
char str1[MAXL],str2[MAXL];
int extend[MAXL],Next[MAXL];
int str1l,str2l;
//模式串自己和自己匹配
void GetNext()
{
    //a爲已匹配位置之首，p爲已匹配位置之尾
    int a=0,p=0;
    //從下標爲0開始的匹配就是自己的長度
    Next[0]=str2l;
    //從下標爲1開始匹配
    for(int i=1;i<str2l;i++)
    {
        //i>=p作用：舉個例子，母串無一字符與模式串相同
        //i大於p或可以繼續往後匹配
        if(i>=p||i+Next[i-a]>=p)
        {
            if(i>=p)
            {
                //更新尾位置
                p=i;
            }
            //往後匹配並更新尾位置
            while(p<str2l&&str2[p]==str2[p-i])
            {
                p++;
            }
            Next[i]=p-i;
            //更新首位置
            a=i;
        }
        else
        {
            Next[i]=Next[i-a];
        }
    }
}
//母串和模式串匹配,與求Next過程類似
void GetExtend()
{
    int a=0,p=0;
    GetNext();
    for(int i=0;i<str1l;i++)
    {
        if(i>=p||i+Next[i-a]>=p)
        {
            if(i>=p)
            {
                p=i;
            }
            while(p<str1l&&p-i<str2l&&str1[p]==str2[p-i])
            {
                p++;
            }
            extend[i]=p-i;
            a=i;
        }
        else
        {
            extend[i]=Next[i-a];
        }
    }
}

int main()
{
    //freopen("in.txt","r",stdin);
    int n;
    scanf("%d",&n);
    getchar();
    while(n--)
    {
        gets(str1);
        str1l=strlen(str1);
        reverse(str1,str1+str1l);
        gets(str2);
        str2l=strlen(str2);
        reverse(str2,str2+str2l);
        GetExtend();
        long long sum=0;
        for(int i=0;i<str1l;i++)
        {
            sum=(sum+((long long)extend[i]*(extend[i]+1)/2)%MOD)%MOD;
        }
        printf("%lld\n",sum);
    }
    return 0;
}