Common Subsequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18387 Accepted Submission(s): 7769
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
Source
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题意：
给你两个字符串。要你找它们的最长公共子串。
思路：
dp[i][j]表示s1长度为i的前缀和s2长度为j的前缀的最长公共子序列的长度。
考虑
s1的第i+1个字符位置要么匹配。要么不匹配。匹配的话只能和前j个位置匹配。
如果s1[i+1]==s2[j]的话
那么dp[i+1][j]=dp[i][j-1]+1很显然成立。
如果s1[i+1]!=s2[j]的话
i+1只能和前s2 的前j-1个位置匹配。
dp[i+1][j]=dp[i+1][j-1]
不匹配前s1的i+1个位置的话。
dp[i+1][j]=dp[i][j]。
那么dp[i+1][j]=max(dp[i][j],dp[i+1][j-1])
这样就把s1的第i+1个位置确定了。
详细见代码：
#include <iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
const int maxn=1010;
int dp[maxn][maxn];
char s1[maxn],s2[maxn];
int main()
{
int i,j,len1,len2;
while(~scanf("%s%s",s1+1,s2+1))
{
len1=strlen(s1+1);
len2=strlen(s2+1);
memset(dp[0],0,sizeof dp[0]);
for(i=1;i<=len1;i++)
{
for(j=1;j<=len2;j++)
{
if(s1[i]==s2[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
",dp[len1][len2]);
}
return 0;
}