Common Subsequence hdu
来源:互联网 发布:搜达数据 编辑:程序博客网 时间:2024/06/08 04:05
Common Subsequence
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64uDescription
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.
Input
abcfbc abfcabprogramming contest abcd mnp
Output
420
Sample Input
abcfbc abfcabprogramming contest abcd mnp
Sample Output
420
最长子序列,大水题
#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
int dp[1000][1000];
int main()
{
char s[1000],s1[1000];
while(~scanf("%s%s",s+1,s1+1))
{
int n,n1;
n=strlen(s+1);
n1=strlen(s1+1);
memset(dp,0,sizeof(dp));
for(int i=1;i<=n;i++)
for(int j=1;j<=n1;j++)
{
if(s[i]==s1[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[n][n1]);
}
return 0;
}
0 0
- hdu 1159 Common Subsequence
- HDU 1159 Common Subsequence
- HDU 1159 Common Subsequence
- HDU 1158 Common Subsequence
- hdu 1159 Common Subsequence
- HDU 1159 Common Subsequence
- HDU 1159 Common Subsequence
- hdu 1159 Common Subsequence
- hdu 1008 Common Subsequence
- Common Subsequence HDU dp
- 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-Common Subsequence
- hdu 1159 Common Subsequence
- Spring集成Quartz定时任务
- 用AutoLayout的一些配置
- 一个好用的反编译网站
- GOTURN 100fps 深度回归网络跟踪
- Java 枚举(enum) 详解7种常见的用法
- Common Subsequence hdu
- CSS基础-介绍及语法
- C#中的文件流总结
- java String.getBytes()编码问题
- 所谓的Java的引用传递
- linux中变量$#,$@,$0,$1,$2,$*,$$,$?的含义
- 红黑树(二)
- HQL 取一个表中的一个字段的非重复数值
- 搭建Spark源码阅读环境