HDU1159-Common Subsequence（dp）

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 38486    Accepted Submission(s): 17658

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

最长公共子序列，状态转移方程为dp[i][j]=max(dp[i-1][j],dp[i][j-1])，即每次保存两串中公共子串最长的那部分，如果出现相同就+1；

#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#define inf 0x3f3f3f3f#define mem(a,b) memset(a,b,sizeof(a))using namespace std;int dp[1005][1005];char a[1005];char b[1005];int main(){    while(~scanf("%s %s",a,b))    {        int la=strlen(a);        int lb=strlen(b);             mem(dp,0);        for(int i=1; i<=la; i++)        {            for(int j=1; j<=lb; j++)            {                if(a[i-1]==b[j-1])                    dp[i][j]=dp[i-1][j-1]+1;                else                    dp[i][j]=max(dp[i][j-1],dp[i-1][j]);            }        }        printf("%d\n",dp[la][lb]);    }}