文章标题 POJ 1458 ： Common Subsequence （最长公共子序列）

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本文介绍了一种解决最长公共子序列问题的有效算法。通过动态规划方法确定两个字符串间的最长公共子序列长度，提供了完整的C++实现代码。

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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 abfcab
programming contest
abcd mnp
Sample Output
4
2
0
题意：给出两个字符串，求出这两个串最长的公共串
分析：dp[i][j]表示第一个串中前i个字符与第二个串前j个串中最长的公共字符数目，然后如果a[i]=b[j],dp[i][j]=dp[i-1][j-1]+1,如果不相等，那么dp[i][j]=max(dp[i-1][j],dp[i][j-1)
代码：

#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<vector>
#include<math.h>
#include<map>
#include<queue> 
#include<algorithm>
using namespace std;
const int inf = 0x3f3f3f3f;
string a,b;
int dp[1005][1005];
int main ()
{
    while (cin>>a>>b){
        int len1=a.length();//长度
        int len2=b.length();
        memset (dp,0,sizeof (dp));//初始化
        for (int i=1;i<=len1;i++){
            for (int j=1;j<=len2;j++){
                if (a[i-1]==b[j-1]){
                    dp[i][j]=dp[i-1][j-1]+1;
                }
                else {
                    dp[i][j]=max(dp[i][j-1],dp[i-1][j]);
                }
            }
        }
        cout<<dp[len1][len2]<<endl;
    }   
    return 0;
}