HDOJ 1159(POJ 1458)Common Subsequence （最长公共子序列 LCS）

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 28498    Accepted Submission(s): 12738

Problem 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, xij = 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.
The program input is from a text file. Each data set in the file contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct. 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

注：此题为：HDOJ 1159 Common Subsequence

   POJ 链接：POJ 1458 Common Subsequence

说明：求最大公共子序列长度           LCS       模板

已AC代码：

#include<cstdio>#include<cstring>#define max(x,y) (x>y?x:y)char ch1[1010],ch2[1010];int dp[1010][1010];int main(){while(scanf("%s%s",ch1,ch2)!=EOF){memset(dp,0,sizeof(dp));int len1=strlen(ch1);int len2=strlen(ch2);int i,j;for(i=1;i<=len1;++i)   //模板 {for(j=1;j<=len2;++j){if(ch1[i-1]==ch2[j-1])dp[i][j]=dp[i-1][j-1]+1;elsedp[i][j]=max(dp[i-1][j],dp[i][j-1]);}}printf("%d\n",dp[len1][len2]);}return 0;}