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:

abcdghaedfhr

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

a1b2c3d4ezz1yy2xx3ww4vvabcdghaedfhrabcdefghijklmnopqrstuvwxyza0b0c0d0e0f0g0h0i0j0k0l0m0n0o0p0q0r0s0t0u0v0w0x0y0z0abcdefghijklmnzyxwvutsrqpoopqrstuvwxyzabcdefghijklmn

Output for the sample input

432614 题目大意：求最长公共子序列 设d[i][j]为A1,A2...Ai和B1,B2...Bj的LCS长度，则的d[i][j]=max{d[i-1][j],d[i][j-1]}，如果A[i]=B[j]，d[i][j]=max{d[i][j],d[i-1][j-1]+1}，边界条件是最外层为0 时间复杂度为O(nm),n、m分别为序列A、B的长度

1 # include
         
           2 # include
          
            3 # define maxn 1005        //这样定义max函数，是不是就不用考虑a，b的类型了 4 # define max(a,b) a>b?a:b 5 int dp[maxn][maxn]; 6 char a[maxn],b[maxn]; 7 int main(){ 8     int lena,lenb,i,j; 9     while(gets(a)&&gets(b)){    //题目中这里不能用scanf，只能用gets输入10         lena=strlen(a);11         lenb=strlen(b);12 13         for(i=0;i<=lena;i++)14             for(j=0;j<=lenb;j++)15                 dp[i][j] = 0;16             17             for(i=1;i<=lena;i++)18             {19                 for(j=1;j<=lenb;j++)20                 {21                     dp[i][j] = max(dp[i-1][j],dp[i][j-1]);    22                     if(a[i-1]==b[j-1])                        23                         dp[i][j] = max(dp[i][j] , dp[i-1][j-1] + 1);24                 }            25             }    26             27             printf("%d\n",dp[lena][lenb]);28     }29     return 0;30 }