LCS POJ 1458 Common Subsequence

题目传送门

题意：输出两字符串的最长公共子序列长度

分析：LCS(Longest Common Subsequence)裸题。状态转移方程：dp[i+1][j+1] = dp[i][j] + 1; (s[i] == t[i])dp[i+1][j+1] = max (dp[i][j+1], dp[i+1][j]); (s[i] != t[i])

代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;const int MAXN = 3e2 + 10;
const int INF = 0x3f3f3f3f;char s[MAXN];
char t[MAXN];
int dp[MAXN][MAXN];void LCS(int len1, int len2)
{
    for (int i=0; i<len1; ++i)
    {
        for (int j=0; j<len2; ++j)
        {
            if (s[i] == t[j])
            {
                dp[i+1][j+1] = dp[i][j] + 1;
            }
            else
            {
                dp[i+1][j+1] = max (dp[i][j+1], dp[i+1][j]);
            }
        }
    }
}int main(void)      //POJ 1458 Common Subsequence
{
    #ifndef ONLINE_JUDGE
    freopen ("LCS.in", "r", stdin);
    #endif    while (~scanf ("%s%s", &s, &t))
    {
        memset (dp, 0, sizeof (dp));
        int len1 = strlen (s);
        int len2 = strlen (t);        LCS (len1, len2);
        printf ("%d\n", dp[len1][len2]);
    }    return 0;
}