Longest Common Subsequence

Source

https://onlinejudge.u-aizu.ac.jp/problems/ALDS1_10_C

Description

Time Limit : 1 sec , Memory Limit : 131072 KB 

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

#include<stdio.h>
#include<string.h>
#define M 1001
void Print_lCS(char x[],char y[]);
int main()
{
    char x[M],y[M];
    int q;
    scanf("%d",&q);
    while(q--){
        scanf("%s%s",x,y);
        Print_lCS(x,y);
    }
}

void Print_lCS(char x[],char y[])
{
    int nx = strlen(x);
    int ny = strlen(y);
    int i,j;
    int c[nx+1][ny+1];
    //边界条件
    for(i = 0;i <= nx;i++)
        c[i][0] = 0;
    for(j = 0;j <= ny;j++)
        c[0][j] = 0;
    //
    for(i = 1;i <= nx;i++){
    for(j = 1;j <= ny;j++){
        if(x[i-1] == y[j-1])
                c[i][j] = c[i-1][j-1] + 1;
            else if(c[i][j-1] >= c[i-1][j])
                c[i][j] = c[i][j-1];
            else
                c[i][j] = c[i-1][j];
        }
    }
    printf("%d
",c[nx][ny]);
}