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

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;
}