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：

dp[0][j] = 0; (0<=j<=m)

dp[i][0] = 0; (0<=i<=n)

dp[i][j] = dp[i-1][j-1] +1; (str1[i] == str2[j])

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);
}