首页 > 代码库 > HDU 1159 Common Subsequence (LCS)
HDU 1159 Common Subsequence (LCS)
Common Subsequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24972 Accepted Submission(s): 11071
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.
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
Recommend
Ignatius
这是一道LCS的裸题：求两个字符串的最长的公共子序列
那么可以定义状态：dp[i][j]以a字符串的第i个字符结尾并且以b字符串的第j个字符结尾的最长公共子序列的长度
状态转移：
if(a[i]==b[j])，dp[i][j]=dp[i-1][j-1]+1; //当两个字符串的中的某一个字符相同时，状态由当前字符串前一位的状态转移过来
else dp[i][j]=max(dp[i-1][j],dp[i][j-1])
最后要注意边界以及初始化的问题
dp[i][0]=0;
dp[0][j]=0;
#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<stdlib.h>#include<queue>#include<stack>#include<algorithm>#define LL __int64using namespace std;const int MAXN=1000+5;const int INF=0x3f3f3f3f;const double EPS=1e-9;int dir4[][2]={{0,1},{1,0},{0,-1},{-1,0}};int dir8[][2]={{0,1},{1,1},{1,0},{1,-1},{0,-1},{-1,-1},{-1,0},{-1,1}};int dir_8[][2]={{-2,1},{-1,2},{1,2},{2,1},{2,-1},{1,-2},{-1,-2},{-2,-1}};int dp[1000][1000];char a[1000],b[1000];int main(){ while(scanf("%s %s",a+1,b+1)!=EOF) { int lena=strlen(a+1),lenb=strlen(b+1); int i,j; memset(dp,0,sizeof(dp)); for(i=0;i<=lena;i++) dp[i][0]=0; for(j=0;j<=lenb;j++) dp[0][j]=0; for(i=1;i<=lena;i++) { for(j=1;j<=lenb;j++) { if(a[i]==b[j]) 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[lena][lenb]); } return 0;}
HDU 1159 Common Subsequence (LCS)
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。