hdu 1159 最长公共子序列
来源:互联网 发布:advanced mac cleaner 编辑:程序博客网 时间:2024/05/21 07:03
Common Subsequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 32410 Accepted Submission(s): 14659
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
#include <bits/stdc++.h>using namespace std;char s1[1001], s2[1001];int dp[1001];int main(){ while(scanf("%s%s",s1,s2) !=EOF){ memset(dp,0,sizeof(dp)); int len1 = strlen(s1); int len2 = strlen(s2); int old; int tmp ; for(int i=0;i!=len1;++i){ old = 0; for(int j=0;j!=len2;++j){ tmp = dp[j]; if(s1[i] == s2[j]) dp[j] = old+1; else dp[j] = max(dp[j],dp[j-1]); old = tmp; } } printf("%d\n",dp[len2-1]); }return 0;}
0 0
- hdu 1159 (最长公共子序列)
- hdu 1159 最长公共子序列
- HDU 1159 最长公共子序列
- HDU 1159 最长公共子序列
- hdu 1159 最长公共子序列
- HDU 1159 最长公共子序列
- HDU-1159(DP_最长公共子序列)
- hdu 1159 最长公共子序列
- HDU 1159 最长公共子序列
- hdu 1159 最长公共子序列
- hdu 1159 最长公共子序列
- HDU 1159(最长公共子序列)
- hdu 1159 最长公共子序列问题
- hdu 1159(最长公共子序列)
- hdu 1159(最长公共子序列)
- hdu-1159【最长公共子序列】
- hdu 1159 最长公共子序列
- hdu 1159 最长公共子序列
- 响应式布局设置--@media only screen and (转载)
- 汇编语言学习第十二章-内中断
- uva10328(递推)
- tableView plain样式和group样式区别
- Jupyter Notebook的快捷键
- hdu 1159 最长公共子序列
- EventBus的使用详解
- nyoj 737 合并石子一(dp)
- 项目导入eclipse中无法启动虚拟器
- Jlink20P接口定义
- @RequestMapping
- io流写入操作日记记录中换行需要注意的点
- MJRefresh的那些坑
- 最优二分查找树