05_最长公共子序列问题(LCS)
2024-10-08 03:30:42
问题来源:刘汝佳《算法竞赛入门经典--训练指南》 P60 问题7:
问题描述:给两个子序列A和B,求长度最大的公共子序列。比如1,5,2,6,8,和2,3,5,6,9,8,4的最长公共子序列为5,6,8另一个解是2,6,8)。
分析:设dp[i][j]为A1,A2,...,Ai和B1,B2,...,Bn的LCS长度,则状态转移方程为:
if(A[i]==B[i])
d[i][j] = d[i-][j-]+;
else
d[i][j] = Max{d[i-][j],d[i][j-]};
时间复杂度为O(n*m);其中空间可用滚动数组优化。
例题链接:http://acm.hdu.edu.cn/showproblem.php?pid=1159
例题:hdu 1159
Common Subsequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25784 Accepted Submission(s): 11428
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 abfcab
programming contest
abcd mnp
Sample Output
4
2
0
题意:给两个字符串,求两字符串的LCS
代码:
#include "stdio.h"
#include "string.h"
#define N 1005 int dp[N];
char s1[N],s2[N]; int inline Max(int a,int b) { return a>b?a:b; } int main()
{
int i,j;
int pre,next;
int len1,len2;
while(scanf("%s %s",s1,s2)!=EOF)
{
len1 = strlen(s1);
len2 = strlen(s2);
memset(dp,,sizeof(dp));
for(i=; i<len1; i++)
{
next = ;
for(j=; j<len2; j++)
{
pre = dp[j]; //pre和next将下一次要用到的dp[i-1][j-1]先存起来,实现空间压缩
if(s1[i]==s2[j])
dp[j] = next+;
else
dp[j] = Max(dp[j-],dp[j]);
next = pre;
}
}
printf("%d\n",dp[len2-]);
}
return ;
}
最新文章
- [Bug] 解决透明 Activity 在 Android 6.0 背景不透明
- Evolution项目(1)
- iOS开发——多线程篇——NSThread
- mysql 64 zip download
- Oracle数据库——基本操作
- Ext.Ajax中scope的作用
- python numpy sum函数用法
- C++检测一个文件是否存在
- shell 练习
- 关于IN-LIST迭代
- Cloudera Search配置
- Tip.It诞生记
- linux命令之mv
- pig脚本的参数传入,多个参数传入
- 双飞翼布局的改造 box-sizing和margin负值的应用
- GridView(多选功能)
- 对象的API
- 吴恩达机器学习笔记35-诊断偏差和方差(Diagnosing Bias vs. Variance)
- js数组根据指定字段(true or false)排序
- ASP.NET微信公众号获取AccessToken