POJ 1458 Common Subsequence（LCS）

Common Subsequence

 

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

裸题，不解释，不好意思我又刷博客了。。。

1)最长公共子序列的长度的动态规划方程

　　　　设有字符串a[0...n]，b[0...m]，下面就是递推公式。字符串a对应的是二维数组num的行，字符串b对应的是二维数组num的列。

　　　　

　　　　另外，采用二维数组flag来记录下标i和j的走向。数字"1"表示，斜向下；数字"2"表示，水平向右；数字"3"表示，竖直向下。这样便于以后的求解最长公共子序列。

#include <iostream>#include <stdio.h>#include <cstring>using namespace std;int main(){    int i,j,dp[2][10086],t;    char a[10086],b[10086];    bool now,pre;    while(~scanf("%s%s",a,b))    {        memset(dp,0,sizeof(dp));        int lena=strlen(a),lenb=strlen(b);        for(now=1,pre=0,i=0; i<lena; i++)            for(swap(now,pre),j=0; j<lenb; j++)                if(a[i]==b[j])                    dp[now][j+1]=dp[pre][j]+1;                else                    dp[now][j+1]=dp[pre][j+1]>dp[now][j]?dp[pre][j+1]:dp[now][j];        printf("%d\n",dp[now][lenb]);    }    return 0;}