2017中国大学生程序设计竞赛 - 网络选拔赛 1004 HDU 6153 A Secret (字符串处理 KMP)

题目链接

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 = （33+32+4*1）%1000000007.

题意：

给定两个字符串，一个匹配串一个模式串，我们的目的是在匹配串中找出所有的能够以模式串为前缀的串的长度，将长度累加求得最终的长度。

分析：
这道题最核心的思想就是模式匹配，但是又有一点点的改进。

我们之大模式匹配在找到一个匹配串时候就直接逃过了，这样的话会减少我们中间查找的匹配串，所以我们将模式串中每一个字符上一次出现的位置记录下来，这样的话就能很好的解决这个问题。

记录模式串的每个前缀(即反转前的后缀)出现的次数，考虑kmp的next数组，f[i]代表1_i包含1f[i]，那么ans[f[i]]+=ans[i];

代码：

#include <cstdio>
#include <iostream>
#include<string.h>
#include <algorithm>
#define ll long long
using namespace std;
const int N=1e6+10;
const int MOD=1e9+7;
char a[N],b[N];
int f[N];
ll ans[N];
void get_next()///获取相同的字符上一次出现的位置
{
    int m=strlen(b);
    f[0]=f[1]=0;
    for(int i=1; i<m; i++)
    {
        int j=f[i];
        while(j&&b[i]!=b[j]) j=f[j];
        f[i+1]=b[i]==b[j]?j+1:0;
    }
}
void kmp_count()///模式匹配
{
    int n=strlen(a),m=strlen(b);
    int j=0;
    for(int i=0; i<n; i++)
    {
        while(j&&b[j]!=a[i]) j=f[j];
        if(b[j]==a[i]) j++;
        ans[j]++;
        if(j==m)
        {
            j=f[j];
            if(j>m) j=0;
        }
    }
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s%s",a,b);
        reverse(b,b+strlen(b));
        reverse(a,a+strlen(a));
        memset(ans,0,sizeof(ans));
        get_next();
        kmp_count();
        int m=strlen(b);
        for(int i=m; i>=1; i--)
            ans[f[i]]+=ans[i];
        ll sum=0;
        for(ll i=1; i<=m; i++)
        {
            sum=(sum+i*ans[i])%MOD;
        }
        printf("%lld\n",sum);
    }
    return 0;
}