hdu 1159 Common Subsequence 【LCS 基础入门】
链接:
Common Subsequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17621 Accepted Submission(s): 7401
..., 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.
abcfbc abfcab
programming contest
abcd mnp
4
2
0
题意:
分析:
/**
求长度为 len1 的序列 A 和长度为 len2 的序列 B 的LCS
注意:序列下标从 0 开始
*/
void LCS(int len1,int len2)
{
for(int i = 1; i <= len1; i++)
{
for(int j = 1; j <= len2; j++)
{
if(s1[i-1] == s2[j-1]) dp[i][j] = dp[i-1][j-1]+1;
else
{
int m1 = dp[i-1][j];
int m2 = dp[i][j-1];
dp[i][j] = max(m1, m2);
}
}
}
}
void LCS(int len1,int len2)
{
for(int i = 1; i <= len1; i++)
{
for(int j = 1; j <= len2; j++)
{
if(s1[i-1] == s2[j-1]) dp[i%2][j] = dp[(i-1)%2][j-1]+1;
else
{
int m1 = dp[(i-1)%2][j];
int m2 = dp[i%2][j-1];
dp[i%2][j] = max(m1, m2);
}
}
}
}
code:
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std; const int maxn = 500+10;
int dp[maxn][maxn];
char s1[maxn], s2[maxn]; /**
求长度为 len1 的序列 A 和长度为 len2 的序列 B 的LCS
注意:序列下标从 0 开始
*/
void LCS(int len1,int len2)
{
for(int i = 1; i <= len1; i++)
{
for(int j = 1; j <= len2; j++)
{
if(s1[i-1] == s2[j-1]) dp[i][j] = dp[i-1][j-1]+1;
else
{
int m1 = dp[i-1][j];
int m2 = dp[i][j-1];
dp[i][j] = max(m1, m2);
}
}
}
} int main()
{
while(scanf("%s%s", s1,s2) != EOF)
{
int len1 = strlen(s1);
int len2 = strlen(s2);
memset(dp,0,sizeof(dp)); LCS(len1, len2);
printf("%d\n", dp[len1][len2]);
}
}
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std; const int maxn = 500+10;
int dp[2][maxn];
char s1[maxn], s2[maxn]; void LCS(int len1,int len2)
{
for(int i = 1; i <= len1; i++)
{
for(int j = 1; j <= len2; j++)
{
if(s1[i-1] == s2[j-1]) dp[i%2][j] = dp[(i-1)%2][j-1]+1;
else
{
int m1 = dp[(i-1)%2][j];
int m2 = dp[i%2][j-1];
dp[i%2][j] = max(m1, m2);
}
}
}
} int main()
{
while(scanf("%s%s", s1,s2) != EOF)
{
int len1 = strlen(s1);
int len2 = strlen(s2);
memset(dp,0,sizeof(dp)); LCS(len1, len2);
printf("%d\n", dp[len1%2][len2]);
}
}
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