Common Subsequence

点击打开链接

A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = < x1, x2, ..., xm > another sequence Z = < z1, z2, ..., zk > is a subsequence of X if there exists a strictly increasing sequence < i1, i2, ..., ik > of indices of X such that for all j = 1,2,...,k, x _ij = zj. For example, Z = < a, b, f, c > is a subsequence of X = < a, b, c, f, b, c > with index sequence < 1, 2, 4, 6 >. Given two sequences X and Y the problem is to find the length of the maximum-length common subsequence of X and Y.

Input

The program input is from the std input. Each data set in the input contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct.

Output

For each set of data the program prints on the standard output the length of the maximum-length common subsequence from the beginning of a separate line.

Sample Input

abcfbc         abfcabprogramming    contest abcd           mnp

Sample Output

420

这是一个最长公共子序列的LCS(博客里有）

dp[i][j]         第一个字符串在第i个字符前且第二个串在第j个字符前可构成的最长子序列的长度 

                                                0   i=0 || j=0
dp[i][j] = dp[i-1][j-1]+1    str1[i]==str2[j]
                max(dp[i-1][j],dp[i][j-1])str1[i]!=str2[j]
                                 

本题也是LCS的模板；

代码：

#include<cstdio>#include<cstring>#include<stack>#include<algorithm>using namespace std;char str1[1005];char str2[1005];int dp[1005][1005];int main(){ while( scanf( "%s%s", str1+1, str2+1 ) != EOF ){str1[0] = str2[0] = '0';int l1 = strlen(str1)-1;int l2 = strlen(str2)-1;   memset( dp, 0, sizeof(dp) );for (int i = 1 ; i <= l1 ; i++){                                           for (int j = 1 ; j <= l2 ; j++){                                             if (str1[i] == str2[j])dp[i][j] = dp[i-1][j-1] + 1;elsedp[i][j] = max(dp[i-1][j],dp[i][j-1]);}}printf ("%d\n",dp[l1][l2]);}return 0;}