POJ 1458 最大公共子序列(动态规划练习）

Common Subsequence Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 59136 Accepted: 24644

Description

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

Source

Southeastern Europe 2003

建立状态maxLen[i][j]表示第一个字符串前i个字符字串与第二个字符串前j个字符子串的最大公共子列长度，转移关系由见代码中，可有二维表表示出来

#include<iostream> #include<cstdio> #include<string.h> using namespace std; char sz1[1000]; char sz2[1000]; int maxLen[1000][1000]; int main() { while(cin>>sz1>>sz2) { int length1 = strlen(sz1); int length2 = strlen(sz2); int i,j; for(i=0;i<=length1;i++) maxLen[i][0] = 0; for(j=0;j<length2;j++) maxLen[0][j] = 0; for(i=1;i<=length1;i++) { for(j=1;j<=length2;j++) { if(sz1[i-1]==sz2[j-1]) maxLen[i][j] = maxLen[i-1][j-1] + 1; else maxLen[i][j] = max(maxLen[i-1][j],maxLen[i][j-1]); } } cout<<maxLen[length1][length2]<<endl; } }