算法思想： 假設X = (x1, x2, ..., xm), Y = {y1, y2, ..., yn)的最長公共子序列爲Z = (z1. z2, ..., zk). 則： 1.若xm = yn, 則zk = xm = yn,且Zk-1是Xm-1和Yn-1的最長公共子序列； 2.若xm != yn, 且zk != xm, 則Z是Xm-1和Y的最長公共子序列； 3.若xm != yn, 且zk != yn, 則Z是X和Yn-1的最長公共子序列。 其中：Xm-1 = (x1, x2, ..., xm-1), Yn-1 = (y1, y2, ..., yn-1), Zk-1 = (z1, z2, ..., zk-1). c[i][j]存儲的是Xi和Yj的最長公共子序列的長度，b[i][j]存儲的是c[i][j]的值是由哪一個子問題（如上3個子問題）的解得到的。 |
#include <stdio.h>
#define MAX 100
char x[MAX+1], y[MAX+1];
int b[MAX+1][MAX+1], c[MAX+1][MAX+1], m, n;
static void init_xy(void);
static void lcs_length(void);
static void lcs(int i, int j);
int main(int argc, char **argv)
{
init_xy();
lcs_length();
printf("The lowest common subsequence is: ");
lcs(m, n);
printf("/n");
return 0;
}
static void init_xy(void)
{
int i;
printf("Please input two sequences numbers: ");
scanf("%d %d", &m ,&n);
getchar();
printf("Please input two sequences:/n");
for(i = 1; i <= m; i++){
scanf("%c", &x[i]);
}
getchar();
for(i = 1; i <= n; i++){
scanf("%c", &y[i]);
}
getchar();
}
static void lcs_length(void)
{
int i, j;
for(i = 1; i <= m; i++){
for(j = 1; j <= n; j++){
if(x[i] == y[j]){
c[i][j] = c[i-1][j-1] + 1;
b[i][j] = 1;
}
else if(c[i-1][j] >= c[i][j-1]){
c[i][j] = c[i-1][j];
b[i][j] = 2;
}
else{
c[i][j] = c[i][j-1];
b[i][j] = 3;
}
}
}
printf("The lowest common subsequence lenght = %d/n", c[m][n]);
}
static void lcs(int i, int j)
{
if(i == 0 || j == 0){
return;
}
if(b[i][j] == 1){
lcs(i-1, j-1);
printf("%c ", x[i]);
}
else if(b[i][j] == 2){
lcs(i-1, j);
}
else{
lcs(i, j-1);
}
}
執行結果：
[xxxx@localhost suanfa]$ ./a.out
Please input two sequences numbers: 7 6
Please input two sequences:
ABCBDAB
BDCABA
The lowest common subsequence lenght = 4
The lowest common subsequence is: B C B A
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