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Common Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 40614    Accepted Submission(s): 18740

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

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题意：求两个字符串的最长相同子序列

c代码如下

/*LCSBDCABAABCBDABdp[1][2] = 1dp[1][1] = 0dp[2][1] = 0//子串：连续//子序列：可以不连续 // LCSdp[i][j]//第一个字符串在第i个字符前且第二个串在第j个字符前可构成的最长子序列的长度 dp[i][j] = 0  i=0 || j=0dp[i-1][j-1]+1     str1[i]==str2[j]max(dp[i-1][j],dp[i][j-1]) str1[i]!=str2[j]*/#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int main(){char c1[505];char c2[505];int i,j,l1,l2;while(~scanf("%s%s",c1+1,c2+1)){int a[505][505]={0};c1[0] = c2[0] = '0';l1=strlen(c1)-1;l2=strlen(c2)-1;for(i=1;i<=l1;i++)for(j=1;j<=l2;j++){if(c1[i]==c2[j])a[i][j]=a[i-1][j-1]+1;elsea[i][j]=max(a[i-1][j],a[i][j-1]);}printf("%d\n",a[l1][l2]);}return 0;}