YTU 2402: Common Subsequence

2402: Common Subsequence

时间限制: 1 Sec  内存限制: 32 MB
提交: 63  解决: 33

题目描述

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. 

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输入

abcfbc abfcab
programming contest 
abcd mnp

输出

4
2
0

样例输入

abcfbc abfcabprogramming contest abcd mnp

样例输出

420

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#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;char s1[1000],s2[1000];int dp[1000][1000];int len1,len2;void LCS(){    int i,j;    memset(dp,0,sizeof(dp));    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;            else                dp[i][j] = max(dp[i-1][j],dp[i][j-1]);        }    }}int main(){    while(~scanf("%s%s",s1,s2))    {        len1 = strlen(s1);        len2 = strlen(s2);        LCS();        printf("%d\n",dp[len1][len2]);    }    return 0;}