UVA_10405 Longest Common Subsequence(DP)

Longest Common Subsequence

Given two sequences of characters, print the length of the longest common subsequence of both sequences. For example, the longest common subsequence of the following two sequences:

abcdgh
aedfhr

is adh of length 3.
Input consists of pairs of lines. The first line of a pair contains the first string and the second line contains the second string. Each string is on a separate line and consists of at most 1,000 characters
For each subsequent pair of input lines, output a line containing one integer number which satisfies the criteria stated above.

Sample input

a1b2c3d4e
zz1yy2xx3ww4vv
abcdgh
aedfhrabcdefghijklmnopqrstuvwxyz
a0b0c0d0e0f0g0h0i0j0k0l0m0n0o0p0q0r0s0t0u0v0w0x0y0z0
abcdefghijklmnzyxwvutsrqpo
opqrstuvwxyzabcdefghijklmn

Output for the sample input

4
3
26
14

题意：给出两字符串(可能含有空格)，求出两字符串最长的公共子串。

题解：DP，原理很清楚，写出递归式，注意输入可能含有空格。#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <cstring>
#define MAX_N 1005

using namespace std;

int n,m;
char s[MAX_N],t[MAX_N];
int dp[MAX_N][MAX_N];
int main()
{
while( gets(s) && gets(t) )
{
n = strlen(s);
m = strlen(t);
for( int i = 0; i < n; i++ )
{
for( int j = 0; j < m; j++ )
{
if( s[i] == t[j] )
dp[i+1][j+1] = dp[i][j] + 1;
else
dp[i+1][j+1] = max(dp[i][j+1],dp[i+1][j]);
}
}
cout<<dp
[m]<<endl;
}
return 0;
}