HDU-1159 Common Subsequence (线性dp 最长公共子串)

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 41796    Accepted Submission(s): 19286

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

#include <bits/stdc++.h>using namespace std;char s1[1001], s2[1001];int dp[1001][1001];int main(){    while(scanf("%s %s", s1, s2) != EOF){        memset(dp, 0, sizeof(dp));        int l1 = strlen(s1), l2 = strlen(s2);        for(int i = 0; i < l1; ++i){            for(int j = 0; j < l2; ++j){                if(s1[i] == s2[j]){                    dp[i + 1][j + 1] = dp[i][j] + 1;                }                else{                    dp[i + 1][j + 1] = max(dp[i + 1][j], dp[i][j + 1]);                }            }        }        printf("%d\n", dp[l1][l2]);    }}/*题意：给两个字符串，求最长公共子串。思路：线性dp，dp[i][j]表示第一个子串前i个部分和第二个子串前j个部分的最长公共子串，如果s[i] == s[j]，显然有dp[i][j] = dp[i - 1][j - 1] + 1;如果不想等，那么就是dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])。*/