动态规划 - 最长公共子串

最长公共子串和最长公共子序列在状态转移方程有些类似的地方，不同的是长公共子串要求必须在原串中是连续的，所以一但某处出现不匹配的情况，此处的值就重置为0。

下面给出最长公共子串的状态转移方程：

1. dp[0][j] = 0; (0<=j<=m)
2. dp[i][0] = 0; (0<=i<=n)
3. dp[i][j] = dp[i-1][j-1] +1; (str1[i] == str2[j])
4. dp[i][j] = 0; (str1[i] != str2[j])

不多说上代码---实现了打印最长公共子串的功能，很简单！

#include "stdafx.h"
#include <stdio.h>
#include <string>
#include <iostream>
using namespace std;

#define MAXSIZE 100
char str1[MAXSIZE];
char str2[MAXSIZE];

int dp[MAXSIZE][MAXSIZE];
//'y'代表str1[i] = str2[j];'n'反之
char path[MAXSIZE][MAXSIZE];

void printComStr(int i, int j)
{
    if (path[i][j] == 'n' || i == 0 || j == 0)
        return;
    if (path[i][j] == 'y')
    {
        printComStr(i - 1, j - 1);
        cout << str1[i - 1];
    }


}

int main()
{
    int n, m;
    int indexi, indexj;
    int ans = 0;
    cin >> str1 >> str2;
    n = strlen(str1);
    m = strlen(str2);
    for (int i = 0; i <= n;i++)
    for (int j = 0; j <= m; j++)
    {
        dp[i][j] = 0;
    }
    for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++)
    {
        if (str1[i - 1] == str2[j - 1])
        {
            dp[i][j] = dp[i - 1][j - 1] + 1;
            path[i][j] = 'y';
        }
        else
        {
            dp[i][j] = 0;
            path[i][j] = 'n';
        }

    }

    for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++)
    {
        if (ans < dp[i][j])
        {
            ans = dp[i][j];
            indexi = i;
            indexj = j;
        }
    }
    cout << ans << endl; 
    cout << indexi << ' ' << indexj << endl;
    printComStr(indexi, indexj);
}