HDU 1159-Common Subsequence(LCS 最长公共子序列)
来源:互联网 发布:linux移动目录 编辑:程序博客网 时间:2024/06/05 05:05
Common Subsequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 32746 Accepted Submission(s): 14818
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 | We have carefully selected several similar problems for you: 1087 1176 1003 1058 1069
题目意思:
给两个字符串,求这两个字符串相同字符的个数。解题思路:
#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>using namespace std;#define MAXN 10010int dp[MAXN][MAXN];int main(){ char a[MAXN],b[MAXN]; while(cin>>a>>b) { int i,j; int n=strlen(a),m=strlen(b); for(i=0; i<n; ++i) for(j=0; j<m; ++j) if(a[i]==b[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]); cout<<dp[n][m]<<endl; } return 0;}
#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>using namespace std;#define MAXN 10010int dp[MAXN][MAXN];int Max(int a,int b,int c){ if(a>=b&&a>=c) return a; else if(b>=a&&b>=c) return b; else if(c>=a&&c>=b) return c;}int main(){ char a[MAXN],b[MAXN]; while(cin>>a>>b) { int i,j; int n=strlen(a),m=strlen(b); for(i=0; i<n; ++i) for(j=0; j<m; ++j) if(a[i]==b[j]) dp[i+1][j+1]=Max(dp[i][j]+1,dp[i][j+1],dp[i+1][j]); else dp[i+1][j+1]=max(dp[i][j+1],dp[i+1][j]); cout<<dp[n][m]<<endl; } return 0;}
0 0
- hdu 1159 Common Subsequence(LCS最长公共子序列)
- HDU 1159-Common Subsequence(LCS 最长公共子序列)
- 【HDU 1159】Common Subsequence(最长公共序列LCS算法)
- HDU 1159 & POJ 1458 Common Subsequence(LCS 最长公共子序列O(nlogn))
- HDU/HDOJ 1159/POJ 1458 Common Subsequence(最长公共子序列LCS,滚动数组)
- HDU 1159 Common Subsequence(最长公共子序列(LCS) 动态规划(DP))
- HDU Common Subsequence 最长公共子序列
- HDU Common Subsequence(最长公共子序列)
- 最长公共子序列(Longest Common Subsequence LCS)
- uva 10405 Longest Common Subsequence 最长公共子序列 LCS
- 最长公共子序列(LCS, Longest Common Subsequence), POJ 1458
- PKU-1458 Common Subsequence (最长公共子序列LCS)
- 求最长公共子序列Longest Common Subsequence LCS
- 最长公共子序列(Longest-Common-Subsequence,LCS)
- poj1458--Common Subsequence--最长公共子序列LCS
- 最长公共子序列问题LCS Longest Common Subsequence
- uva10405 - Longest Common Subsequence(LCS,最长公共子序列)
- HD1159 Common Subsequence 最长公共子序列(LCS)
- HTML <iframe> 标签
- bzoj 1003: [ZJOI2006]物流运输(最短路+DP)
- Nginx TCP Load Balance
- Centos 环境搭建 svn
- Mysql主从配置
- HDU 1159-Common Subsequence(LCS 最长公共子序列)
- windows ce 5.0 直接连接局域网内 sql2005数据库
- 探索《How Tomcat Works》心得(三)--容器
- 文件锁
- windows下libfreenect2的安装
- 杂七杂八
- react, Stateless Functions, ES6 花括号参数, Spread operator
- 【bzoj 3531】 [Sdoi2014]旅行(树链剖分+树套树)
- 349. Intersection of Two Arrays【E】