hdoj1159Common Subsequence(LCS 最长公共子序列)

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

 

思路：

两个字符串：s,t，他们的长度分别为n,m;
dp[ i ][ j ]: s1->si和t1->ti对应的LCS的长度.

有下列两种情况：

1， 当s(i+1)==t(j+1)时 有dp[ i +1][ j+1 ]=dp[ i ][ j ]+1;

2,    当s(i+1)!=t(j+1)时  有dp[ i+1][ j+1]=max(dp[ i+1][ j ],dp[ i ][ j+1]);

最后推出dp[ n ][ m ]即为最长公共子序列.

代码如下：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;char str1[1100];char str2[1100];int dp[1100][1100];int main(){while(scanf("%s%s",str1,str2)!=EOF){int len1=strlen(str1);int len2=strlen(str2);int i,j;for(i=0;i<len1;i++)dp[i][0]=0;for(j=0;j<len2;j++)dp[0][j]=0;for(i=0;i<len1;i++){for(j=0;j<len2;j++){if(str1[i]==str2[j])dp[i+1][j+1]=dp[i][j]+1;elsedp[i+1][j+1]=max(dp[i+1][j],dp[i][j+1]);}}printf("%d\n",dp[len1][len2]);}}