URAL 1577. E-mail(簡單二維dp)

畫船聽雨發表於2014-11-02
AI

給你兩個子串，讓你找出來一個最短的字串包含這兩個子串，輸出最多的子串有多少種。

類似於最長公共子序列，相等的話長度+1，不想等的話比較長度，使用長度小的。

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 109 + 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;
}


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