Common Subsequence(dp入门题)

http://acm.hdu.edu.cn/showproblem.php?pid=1159

dp给我的感觉真的很奇妙，以前刚学dfs的时候是定义这个函数有什么功能，现在是定义一个数组有什么功能。

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 29393    Accepted Submission(s): 13215

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

 

水的差不多了，开始专题了

#include <cstdio>#include <algorithm>#include <cstring>using namespace std;char s1[1010],s2[1010];int dp[1010][1010];int main(){    while(scanf("%s%s",s1,s2)==2){        int len1=strlen(s1);        int len2=strlen(s2);        for(int i=0;i<=len1;i++)            dp[i][0]=0;        for(int i=0;i<len2;i++)            dp[0][i]=0;        for(int i=1;i<=len1;i++)            for(int 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]);        printf("%d\n",dp[len1][len2]);    }    return 0;}

AC之路，我选择坚持~~