URAL 1577. E-mail(简单二维dp)

给你两个子串，让你找出来一个最短的字符串包括这两个子串，输出最多的子串有多少种。

类似于最长公共子序列，相等的话长度+1，不想等的话比較长度，使用长度小的。

1577. E-mail

Time limit: 1.0 second
Memory limit: 64 MB

Vasya started to use the Internet not so long ago, so he has only two e-mail accounts at two different servers. For each of them he has a password, which is a non-empty string consisting of only lowercase latin letters. Both mail servers accept a string as a password if and only if the real password is its subsequence.

Vasya has a hard time memorizing both passwords, so he would like to come up with a single universal password, which both servers would accept. Vasya can't remember too long passwords, hence he is interested in a universal password of a minimal length. You are to help Vasya to find the number of such passwords.

Input

The input consists of 2 lines, each of them containing the real password for one of the servers. The length of each password doesn't exceed 2000 characters.

Output

Output the number of universal passwords of minimal length modulo 10 9 + 7.

Samples

input	output
b ab	1
abcab cba	4

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-8
#define M 1000100
#define LL long long
//#define LL long long
#define INF 0x3f3f3f
#define PI 3.1415926535898
#define mod 1000000007

const int maxn = 2010;

using namespace std;

LL dp[maxn][maxn];
LL len[maxn][maxn];

char str1[maxn];
char str2[maxn];

int main()
{
    while(~scanf("%s %s",str1+1, str2+1))
    {
        int n = strlen(str1+1);
        int m = strlen(str2+1);
        memset(dp, 0, sizeof(dp));
        memset(len, 0, sizeof(len));
        for(int i = 0; i <= max(n, m); i++)
        {
            dp[i][0] = 1;
            dp[0][i] = 1;
            len[i][0] = i;
            len[0][i] = i;
        }
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= m; j++)
            {
                if(str1[i] == str2[j])
                {
                    dp[i][j] = dp[i-1][j-1];
                    len[i][j] = len[i-1][j-1]+1;
                    continue;
                }
                if(len[i-1][j] > len[i][j-1])
                {
                    dp[i][j] = dp[i][j-1];
                    len[i][j] = len[i][j-1]+1;
                    continue;
                }
                if(len[i][j-1] > len[i-1][j])
                {
                    dp[i][j] = dp[i-1][j];
                    len[i][j] = len[i-1][j]+1;
                    continue;
                }
                len[i][j] = len[i-1][j]+1;
                dp[i][j] += dp[i-1][j]+dp[i][j-1];
                dp[i][j] %= mod;
            }
        }
        cout<<dp[n][m]<<endl;
    }
    return 0;
}