HDU 1159 Common Subsequence 最长公共子序列

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 40923    Accepted Submission(s): 18892

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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求最长公共子序列模板题，时间复杂度O（m*n），可以过

ac代码

#include <cstdio>#include <cmath>#include <cstring>#include <stack>#include <algorithm>#define inf 0x3f3f3f3fusing namespace std;const int maxn=1000+5;char s1[maxn],s2[maxn];int len1,len2;int dp[maxn][maxn];int main(){while(~scanf("%s%s",s1+1,s2+1)){/*printf("s1==%s\ns2==%s\n",s1+1,s2+1);*/len1=strlen(s1+1);len2=strlen(s2+1);memset(dp,0,sizeof(dp));for(int i=1;i<=len1;i++){for(int j=1;j<=len2;j++){/*printf("%c %c\n",s1[i],s2[j]);*/if(s1[i] == s2[j])dp[i][j]=dp[i-1][j-1]+1;elsedp[i][j]=max(dp[i-1][j],dp[i][j-1]);}}printf("%d\n",dp[len1][len2]);} return 0;}