HDU 1159 Common Subsequence
来源:互联网 发布:usb数据传输线 编辑:程序博客网 时间:2024/06/05 02:45
Common Subsequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 36360 Accepted Submission(s): 16647
Total Submission(s): 36360 Accepted Submission(s): 16647
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
题解:求最长公共子序列……
420
题解:求最长公共子序列……
#include<stdio.h> #include<stdlib.h>#include<math.h>#include<string.h>#include<algorithm>#include<iostream>#include<queue>#include<stack>#define mem(v) memset(v,0,sizeof(v))using namespace std;char x[1001],y[1001];int dp[1002][1002];int main(){ while(~scanf("%s%s",x,y)) { mem(dp); for(int i=1; i<=strlen(x); i++) for(int j=1; j<=strlen(y); j++) { if(x[i-1]==y[j-1]) 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[strlen(x)][strlen(y)]); }}
0 0
- hdu 1159 Common Subsequence
- HDU 1159 Common Subsequence
- HDU 1159 Common Subsequence
- hdu 1159 Common Subsequence
- HDU 1159 Common Subsequence
- HDU 1159 Common Subsequence
- hdu 1159 Common Subsequence
- hdu 1159 Common Subsequence
- hdu 1159 Common Subsequence
- Common Subsequence hdu 1159
- HDU 1159 Common Subsequence
- HDU 1159 Common Subsequence
- hdu 1159 Common Subsequence
- hdu 1159 Common Subsequence
- HDU 1159 Common Subsequence
- hdu 1159 Common Subsequence
- hdu 1159 Common Subsequence
- hdu 1159 Common Subsequence
- JAVA编程风格
- exercise 38 列表操作
- 58. Length of Last Word
- ABAP Bom按层展开的几种实现方法
- 139. Word Break
- HDU 1159 Common Subsequence
- 解决android扫描二维码时,用户禁止权限报错问题
- eMMC 原理 :Flash Memory 简介
- canny边缘检测C++实现
- GitLab的安装实战
- 启动顺序导致的 ntldr is missing 错误情况
- 设计模式---十一个行为型模式
- 01分数规划三类型总结
- 米洛个人修炼术:注意这三方面,天天都早起