POJ 1458 Common Subsequence

Common Subsequence

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 30985 Accepted: 12073

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         abfcabprogramming    contest abcd           mnp

Sample Output

420

Source

Southeastern Europe 2003

解题思路：题目说那么多都不用管，就是求两串的最大公共子序列，因为在求的时候，每个状态都已经保证了在第二个序列中的子序列序号是递增的。

dp[i][j]表示第一个串的前i个字母和第二个串的前j个字母配对的最大匹配数，初始化为dp[i][j]={0},

状态转移方程为dp[i][j]={dp[i-1][j-1]+1(若s1[i-1]==s2[j-1]),max(dp[i-1][j],dp[i][j-1])};

#include<iostream>using namespace std;inline int max(int a,int b){return a>b?a:b;}int main(){char s1[500],s2[500];int i,j,len1,len2,dp[500][500];while(cin>>s1>>s2){memset(dp,0,sizeof(dp));        len1=strlen(s1);len2=strlen(s2);    for(i=1;i<=len1;i++)for(j=1;j<=len2;j++){if(s1[i-1]==s2[j-1])dp[i][j]=dp[i-1][j-1]+1;elsedp[i][j]=max(dp[i-1][j],dp[i][j-1]);}cout<<dp[len1][len2]<<endl;}}