uva 10405 - Longest Common Subsequence(注意空格陷阱)

1、http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1346

2、题目大意：给定两个字符串，求出最长的公共子串的长度，挺简单的题目，居然wrong在空格陷阱上了

状态转移方程

        if(str1[i]==str2[j])
                dp[i][j]=dp[i-1][j-1]+1;
        else
                dp[i][j]=max(dp[i-1][j],dp[i][j-1]);

3、题目：

Longest Common Subsequence

Sequence 1:

Sequence 2:


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
aedfhr
abcdefghijklmnopqrstuvwxyz
a0b0c0d0e0f0g0h0i0j0k0l0m0n0o0p0q0r0s0t0u0v0w0x0y0z0
abcdefghijklmnzyxwvutsrqpo
opqrstuvwxyzabcdefghijklmn

Output for the sample input

4
3
26
14

4、ac代码：

#include<stdio.h>
#include<string.h>
#define N 1005
#include<algorithm>
using namespace std;
char str1[N];
char str2[N];
int dp[N][N];
int main()
{
    while(gets(str1+1),gets(str2+1))//注意题目题目可能有空格，用scanf果断的wrong answer了
    {
        memset(dp,0,sizeof(dp));
        int len1=strlen(str1+1);
        int len2=strlen(str2+1);
        for(int i=1;i<=len1;i++)
        {
            for(int j=1;j<=len2;j++)
            {
                if(str1[i]==str2[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[len1][len2]);
    }
    return 0;
}
/*

a1b2c3d4e
zz1yy2xx3ww4vv
abcdgh
aedfhr
abcdefghijklmnopqrstuvwxyz
a0b0c0d0e0f0g0h0i0j0k0l0m0n0o0p0q0r0s0t0u0v0w0x0y0z0
abcdefghijklmnzyxwvutsrqpo
opqrstuvwxyzabcdefghijklmn
*/