杭电 1159 Common Subsequence

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.

Sample Input

abcfbc abfcab

programming contest

abcd mnp

Sample Output

4 2 0

求最长公共子序列

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<string.h>
 4 using namespace std;
 5 int dp[10000][10000];
 6 int main()
 7 {
 8     int lena,lenb;
 9     char a[10000],b[10000];
10     while(scanf("%s %s",&a,&b)!=EOF)
11     {
12         int i,j;
13         lena=strlen(a);
14         lenb=strlen(b);
15         for(i = 1 ; i <= lena ; i++)
16         {
17             for(j = 1 ; j <= lenb ; j++)
18             {
19                 if(a[i-1] == b[j-1])
20                     dp[i][j]=dp[i-1][j-1]+1;
21                 else
22                     dp[i][j]=max(dp[i][j-1],dp[i-1][j]);
23             }
24         }
25         printf("%d\n",dp[lena][lenb]);
26     }
27 }
时间: 2024-11-10 07:04:28

杭电 1159 Common Subsequence的相关文章

杭电1159 Common Subsequence【最长公共子序列】

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1159 解题思路:任意先给出两个字符串 abcfbc abfcab,用dp[i][j]来记录当前最长的子序列,则如果有x[i]与y[j]相等的话,则相当于公共子序列的长度在dp[i-1][j-1]上增加1, 如果x[i]与y[j]不相等的话,那么dp[i][j]就取得dp[i][j-1]和dp[i-1][j]中的最大值即可.时间复杂度为O(mn) 反思:大概思路想出来之后,因为dp数组赋初值调了很久,

杭电1159(Common Subsequence)LCS和dp

点击打开杭电1159 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 ex

POJ 1159 Palindrome &amp;&amp; HDU 1159 Common Subsequence

1.先说说杭电的1159吧! 这道题是基础动规,比较简单! 就是要你求最长的公共子序列(不要优化) 动态转移方程: dp[i+1][j+1]=(a[i]=b[i])?dp[i][j]+1:max(dp[i+1][j],dp[i][j+1]) AC代码: #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; #define N 520 char a[N],b[N]; in

hdu 1159 Common Subsequence(最长公共子序列 DP)

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1159 Common Subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 25416    Accepted Submission(s): 11276 Problem Description A subsequence of

hdu 1159 Common Subsequence(最长公共子序列)

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1159 Common Subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 37551    Accepted Submission(s): 17206 Problem Description A subsequence of

HDU 1159——Common Subsequence(DP)

Common Subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 23279    Accepted Submission(s): 10242 Problem Description A subsequence of a given sequence is the given sequence with some e

HDU 1159 Common Subsequence(裸LCS)

传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1159 Common Subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 47676    Accepted Submission(s): 21890 Problem Description A subsequence of

杭电 HDU ACM 1159 Common Subsequence

Common Subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 26493    Accepted Submission(s): 11771 Problem Description A subsequence of a given sequence is the given sequence with some e

hdu 1159 Common Subsequence(dp 最长公共子序列问题LCS)

最长公共子序列问题(LCS,Longerst Common Subsequence). s1s2……si+1和t1t2……tj+1的公共子序列可能是: ①当si+1=tj+1时,在s1s2……si+1和t1t2……tj+1的公共子序列末尾追加一个. ②s1s2……si+1和t1t2……tj的公共子序列 ③s1s2……si和t1t2……tj+1的公共子序列 所以易得到递推关系dp[i+1][j+1]=  max{ dp[i][j]+1 , dp[i][j+1] , dp[i+1][j]) }