POJ 1458 Common Subsequence（LCS最长公共子序列）

POJ 1458 Common Subsequence（LCS最长公共子序列）解题报告

题目链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87730#problem/F

题目：

Common Subsequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 43388		Accepted: 17613

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

题目大意：
给出两个字符串，求两字符串的最长公共子序列。

分析:
很明显用LCS，时间复杂度O（nm）,其中n,m是序列A和B的长度。当s1[i-1]==s2[k-1]时，d(i,k)=d(i-1,k-1)+1,否则，
d(i,k)=max{d(i-1,k),d(i,k-1)}。

代码：

 #include<cstdio>
 #include<iostream>
 #include<cstring>
 using namespace std;

 const int maxn=;

 int dp[maxn][maxn];

 int max(int a,int b)
 {
     return a>b?a:b;
 }

 int main()
 {
     char s1[maxn],s2[maxn];
     while(scanf("%s%s",s1,s2)!=EOF)
     {
         int m=strlen(s1);
         int n=strlen(s2);
         memset(dp,,sizeof(dp));
         for(int i=;i<=m;i++)
         {
             for(int k=;k<=n;k++)
             {
                 if(s1[i-]==s2[k-])   //s1和s2相等，dp+1
                     dp[i][k]=dp[i-][k-]+;
                 else                   //s1和s2不相等时看下一个
                     dp[i][k]=max(dp[i-][k],dp[i][k-]);
             }
         }
         printf("%d\n",dp[m][n]);
     }
     return ;
 }