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
参考： http://blog.csdn.net/hrn1216/article/details/51534607
题意：求最长公共子序列。子序列，可不相邻，但相等顺序不能改变
思路：动态规划，状态方程为：dp【i】【j】=dp【i-1】【j-1】（a【i】==b【j】）
or dp【i】【j】=max（dp【i-1】【j】，dp【i】【j-1】）;
dp【i】【j】表示a字符串第i个之前，和b字符串第j个之前的最长公共子序列的长度。
其他见参考（实在描述不了，脑补dp生成过程）
感想：动态规划真的难以理解，更难以找到规律啊！！！
AC代码：
#include<iostream>#include<cstdio>#include<cstring>#include<cmath>using namespace std;int main(){ // freopen ("in.txt","r",stdin); char a[500]; char b[500]; while (cin>>a>>b) { int n=strlen(a); int m=strlen(b); int dp[n+5][m+5]; for (int i=0;i<=n;i++) for (int j=0;j<=m;j++) { if (i&&j) { if (a[i-1]==b[j-1]) dp[i][j]=dp[i-1][j-1]+1; else dp[i][j]=max(dp[i-1][j],dp[i][j-1]); } else dp[i][j]=0; } /* for (int i=0;i<=n;i++) for (int j=0;j<=m;j++) { cout << dp[i][j] ; if (j==m) cout <<endl; else cout <<" "; }*/ cout << dp[n][m] <<endl; memset (a,0,sizeof(a)); memset (b,0,sizeof(b)); } return 0;}
