HDU 1159 Common Subsequence 最长公共子序列 详解

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 40051    Accepted Submission(s): 18450

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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最长公共子序列水题，题意不用读了，看实例直接就可以做题。

由于直接用scanf输入的字符串，下标从0开始，但是在dp里面，下标我从1开始，加1即可

ac代码（空间复杂度O(n^2)）

#include <stdio.h> #include <cmath>#include <cstring>#include <queue>#include <algorithm>#define ll long long using namespace std;char s1[1005],s2[1005];int cnt;int n,m;int dp[1005][1005];int main(){while(~scanf("%s%s",s1,s2)){memset(dp,0,sizeof(dp));n=strlen(s1);m=strlen(s2);for(int i=0;i<n;i++){for(int j=0;j<m;j++){if(s1[i]==s2[j]){dp[i+1][j+1]=dp[i][j]+1;}else{dp[i+1][j+1]=max(dp[i][j+1],dp[i+1][j]);}}}printf("%d\n",dp[n][m]);}return 0;}