HDU

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 40191    Accepted Submission(s): 18522

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

 

Source

Southeastern Europe 2003

 

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#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int m,n;char s[1001],t[1001];int dp[1001][1001];//定义dp[i][j]记录序列最长子序列的长度，//合法状态的初始值为当序列X的长度为0或Y的长度为0，公共子序列最长长度为0，//即dp[i][j]=0，所以用i和j分别表示序列s的长度和序列t的长度void solve(){    int i,j;    for(int i=0; i<n; i++)    {        for(int j=0; j<m; j++)        {            if(s[i]==t[j])                dp[i+1][j+1]=dp[i][j]+1;    //若s中的记录序列下一个和t中记录的子序列下一个相等，就把最长子序列个数加一            else                dp[i+1][j+1]=max(dp[i][j+1],dp[i+1][j]); //若不相等，t，s重新从下一个开始记录，找出最长的公共子序列        }    }    printf("%d\n",dp[n][m]);}int main(){    while(~scanf("%s%s",s,t))    {        n=strlen(s);        m=strlen(t);        solve();    }}