Longest Common Subsequence

For given two sequences $\mathrm{XX and YY,\; a\; sequence ZZ is\; a\; common\; subsequence\; of XX and YY if ZZ is\; a\; subsequence\; of\; both XX and YY.\; For\; example,\; if X=\{a,b,c,b,d,a,b\}X=\{a,b,c,b,d,a,b\} and Y=\{b,d,c,a,b,a\}Y=\{b,d,c,a,b,a\},\; the\; sequence \{b,c,a\}\{b,c,a\} is\; a\; common\; subsequence\; of\; both XX and YY.\; On\; the\; other\; hand,\; the\; sequence \{b,c,a\}\{b,c,a\} is\; not\; a\; longest\; common\; subsequence\; (LCS)\; of XX and YY,\; since\; it\; has\; length\; 3\; and\; the\; sequence \{b,c,b,a\}\{b,c,b,a\},\; which\; is\; also\; common\; to\; both XX and YY,\; has\; length\; 4.\; The\; sequence \{b,c,b,a\}\{b,c,b,a\}is\; an\; LCS\; of XX and YY,\; since\; there\; is\; no\; common\; subsequence\; of\; length\; 5\; or\; greater.}$

Write a program which finds the length of LCS of given two sequences $\mathrm{XX and YY.\; The\; sequence\; consists\; of\; alphabetical\; characters.}$

Input

The input consists of multiple datasets. In the first line, an integer $\mathrm{qq which\; is\; the\; number\; of\; datasets\; is\; given.\; In\; the\; following 2\times q2\times q lines,\; each\; dataset\; which\; consists\; of\; the\; two\; sequences XX and YY are\; given.}$

Output

For each dataset, print the length of LCS of $\mathrm{XX and YY in\; a\; line.}$

Constraints

• $\mathrm{1\le q\le 1501\le q\le 150}$
• $\mathrm{1\le 1\le length\; of XX and YY \le 1,000\le 1,000}$
• $\mathrm{q\le 20q\le 20 if\; the\; dataset\; includes\; a\; sequence\; whose\; length\; is\; more\; than\; 100}$

Sample Input 1

3
abcbdab
bdcaba
abc
abc
abc
bc

Sample Output 1

4
3
2

Reference

Introduction to Algorithms, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. The MIT Press.

Code

 1 #include<stdio.h>
 2 #include<string.h>
 3 #define M 1001
 4 void Print_lCS(char x[],char y[]);
 5 int main()
 6 {
 7     char x[M],y[M];
 8     int q;
 9     scanf("%d",&q);
10     while(q--){
11         scanf("%s%s",x,y);
12         Print_lCS(x,y);
13     }
14 }
15 
16 void Print_lCS(char x[],char y[])
17 {
18     int nx = strlen(x);
19     int ny = strlen(y);
20     int i,j;
21     int c[nx+1][ny+1];
22     //边界条件
23     for(i = 0;i <= nx;i++)
24         c[i][0] = 0;
25     for(j = 0;j <= ny;j++)
26         c[0][j] = 0;
27     //
28     for(i = 1;i <= nx;i++){
29     for(j = 1;j <= ny;j++){
30         if(x[i-1] == y[j-1])
31                 c[i][j] = c[i-1][j-1] + 1;
32             else if(c[i][j-1] >= c[i-1][j])
33                 c[i][j] = c[i][j-1];
34             else
35                 c[i][j] = c[i-1][j];
36         }
37     }
38     printf("%d\n",c[nx][ny]);
39 }

c_code