我想打印所有可能的LCS问题解决方案。
两个字符串abcbdab和bdcaba应打印以下3个字符串： bdab，BCBA，bcab。
C是全局矩阵表，根据算法取值，m，n是序列a，b的长度。
但输出是出乎意料的。
#include<stdio.h>
#include<conio.h>
int co=0,m=0,n=0,c[10][10];
char a[10],b[10];
void main()
{
int i,j;
clrscr();
printf("Enter Two strings: ");
scanf("%s",a);
scanf("%s",b);
m=strlen(a);
n=strlen(b);
for(i=0;i<=m;i++)
{
for(j=0;j<=n;j++)
{ if(i==0 || j==0)
{
c[i][j]=0;
}
else if(a[i-1]==b[j-1])
{
c[i][j]=c[i-1][j-1]+1;
}
else if(c[i-1][j]>=c[i][j-1])
{
c[i][j]=c[i-1][j];
}
else
{
c[i][j]=c[i][j-1];
}
}
}
for(i=0;i<=m;i++)
{
for(j=0;j<=n;j++)
{
printf("%d\t",c[i][j]);
}
printf("\n");
}
print(m,n);
getch();
}
print(int i,int j)
{
if(i==0 || j==0)
return 0;
else if(a[i-1]==b[j-1])
{
print(i-1,j-1);
if(co==c[m][n])
{
co=0;
printf("\n");
}
printf("%c",a[i-1]);
co++;
}
else if(c[i-1][j]==c[i][j-1])
{
print(i-1,j);
print(i,j-1);
}
else if(c[i][j-1]>=c[i-1][j])
print(i,j-1);
else
print(i-1,j);
return;
}
答案 0 :(得分：6)
在这里，您可以找到如何执行此操作的递归方法：Reading out all LCSs
以下是我在Java中使用此方法的代码：
private Set<String> lcs(int[][] dp, String fst, String snd, int i, int j) {
Set<String> lcss = new HashSet<>();
if (i == 0 || j == 0) {
lcss.add("");
} else if (fst.charAt(i - 1) == snd.charAt(j - 1)) {
for (String lcs : lcs(dp, fst, snd, i - 1, j - 1)) {
lcss.add(lcs + fst.charAt(i - 1));
}
} else {
if (dp[i - 1][j] >= dp[i][j - 1]) {
lcss.addAll(lcs(dp, fst, snd, i - 1, j));
}
if (dp[i][j - 1] >= dp[i - 1][j]) {
lcss.addAll(lcs(dp, fst, snd, i, j - 1));
}
}
return lcss;
}
答案 1 :(得分：0)
您的源代码未打印lcs。它实际上是在计算lcs的长度。您提供的源代码完全错误。首先尝试打印一个lcs。然后扩展该解决方案以打印所有lcs。下面给出的帮助是java解决方案。
static int arr[][];
static void lcs(String s1, String s2) {
for (int i = 1; i <= s1.length(); i++) {
for (int j = 1; j <= s2.length(); j++) {
if (s1.charAt(i - 1) == s2.charAt(j - 1))
arr[i][j] = arr[i - 1][j - 1] + 1;
else
arr[i][j] = Math.max(arr[i - 1][j], arr[i][j - 1]);
}
}
}
static Set<String> lcs(String s1, String s2, int len1, int len2) {
if (len1 == 0 || len2 == 0) {
Set<String> set = new HashSet<String>();
set.add("");
return set;
}
if (s1.charAt(len1 - 1) == s2.charAt(len2 - 1)) {
Set<String> set = lcs(s1, s2, len1 - 1, len2 - 1);
Set<String> set1 = new HashSet<>();
for (String temp : set) {
temp = temp + s1.charAt(len1 - 1);
set1.add(temp);
}
return set1;
} else {
Set<String> set = new HashSet<>();
Set<String> set1 = new HashSet<>();
if (arr[len1 - 1][len2] >= arr[len1][len2 - 1]) {
set = lcs(s1, s2, len1 - 1, len2);
}
if (arr[len1][len2 - 1] >= arr[len1 - 1][len2]) {
set1 = lcs(s1, s2, len1, len2 - 1);
}
for (String temp : set) {
set1.add(temp);
}
//System.out.println("In lcs" + set1);
return set1;
}
}
public static void main(String[] args) {
String s1 = "abcbdab";
String s2 = "bdcaba ";
arr = new int[s1.length() + 1][s2.length() + 1];
lcs(s1, s2);
System.out.println(lcs(s1, s2, s1.length(), s2.length()));
}
如果字符串的最后一个字符相等，那么它们必须是lcs。如果它们不相等，则lcs将从矩阵的上侧或矩阵的左侧构造，这取决于哪个值更大。如果两个值都相等，则将从两侧构造lcs。所以继续构造lcs，直到你构造了所有的lcs并将它们存储在一个集合中。
答案 2 :(得分：0)
这是带有注释的Java代码，解释了如何打印所有可能的lcs。
import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;
public class LongestCommonSubsequence {
public static int[][] LCSmatrix(String X, String Y) {
//we ignore the top most row and left most column in this matrix
//so we add 1 and create a matrix with appropriate row and column size
int m = X.length() + 1, n = Y.length() + 1;
int[][] c = new int[m][n];
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
//since we added 1 to row size and column size,
// we substract 1 from i,j to find the char at that index
if (X.charAt(i - 1) == Y.charAt(j - 1)) {
c[i][j] = c[i - 1][j - 1] + 1;
} else if (c[i - 1][j] >= c[i][j - 1]) {
c[i][j] = c[i - 1][j];
} else {
c[i][j] = c[i][j - 1];
}
}
}
printMatrix(c);
return c;
}
public static void printMatrix(int[][] grid) {
for (int r = 0; r < grid.length; r++) {
for (int c = 0; c < grid[r].length; c++) {
System.out.print(grid[r][c] + " ");
}
System.out.println();
}
}
public static void allLCS(int[][] c, String X, String Y, int i, int j, Set<String> setLCS, String s) {
//return when either of the string length is 0
if (i == 0 || j == 0) {
setLCS.add(s);
return;
}
//if last characters are equal, they belong in lcs
if (X.charAt(i - 1) == Y.charAt(j - 1)) {
//prepend the char to lcs since, we are going backwards
s = X.charAt(i - 1) + s;
//continue finding lcs in substrings X.substring(0,i-1) and Y.substring(0,j-1)
allLCS(c, X, Y, i - 1, j - 1, setLCS, s);
} // if there is a tie in matrix cells, we backtrack in both ways,
// else one way, which ever is greater
else if (c[i - 1][j] == c[i][j - 1]) {
//continue finding lcs in substring X.substring(0,i-1)
allLCS(c, X, Y, i - 1, j, setLCS, s);
//continue finding lcs in substring Y.substring(0,j-1)
allLCS(c, X, Y, i, j - 1, setLCS, s);
} else if (c[i - 1][j] > c[i][j - 1]) {
allLCS(c, X, Y, i - 1, j, setLCS, s);
} else {
allLCS(c, X, Y, i, j - 1, setLCS, s);
}
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.println(" Enter String X and Y : ");
String X = sc.next();
String Y = sc.next();
sc.close();
Set<String> set = new HashSet<String>();
allLCS(LCSmatrix(X, Y), X, Y, X.length(), Y.length(), set, "");
System.out.println(set.toString());
}
}