原文:
给定源点 (src x 、src y ) 和目的点 (dst x 、dst y ) 的坐标，任务是确定从源点到达目的点的可能路径。从任何一点来看 (x，y) 只有两种有效招式: (2 * x y，y) 或 (x，2 * y x) 。如果路径不存在，则打印 -1 。 例:
输入:【src x 、src 和= { 5.8 }，(dst x 、dst 和= { 83.21 }
方法:想法是使用从每个点 进行两次调用
- 在第一次递归调用中，通过 (2 * x y) 更新‘x’，并且不改变 y 坐标。
- 在第二次递归调用中，通过 (2 * y x) 更新‘y’，并且不改变 x 坐标。
如果当前坐标 x 或 y 超过目的地坐标，我们终止递归调用。将路径存储在中，到达堆栈 的目的打印元素后，执行上述方法:
c
// c program for printing a path from
// given source to destination
#include
using namespace std;
// function to print the path
void printexistpath(stack sx, stack sy, int last)
{
// base condition
if (sx.empty() || sy.empty()) {
return;
}
int x = sx.top();
int y = sy.top();
// pop stores elements
sx.pop();
sy.pop();
// recursive call for printing stack
// in reverse order
printexistpath(sx, sy, last);
if (sx.size() == last - 1) {
cout << "(" << x << ", " << y << ")";
}
else {
cout << "(" << x << ", " << y << ") -> ";
}
}
// function to store the path into
// the stack, if path exist
bool storepath(int srcx, int srcy, int destx, int desty,
stack& sx, stack& sy)
{
// base condition
if (srcx > destx || srcy > desty) {
return false;
}
// push current elements
sx.push(srcx);
sy.push(srcy);
// condition to check whether reach to the
// destination or not
if (srcx == destx && srcy == desty) {
printexistpath(sx, sy, sx.size());
return true;
}
// increment 'x' ordinate of source by (2*x y)
// keeping 'y' constant
if (storepath((2 * srcx) srcy, srcy, destx, desty, sx, sy)) {
return true;
}
// increment 'y' ordinate of source by (2*y x)
// keeping 'x' constant
if (storepath(srcx, (2 * srcy) srcx, destx, desty, sx, sy)) {
return true;
}
// pop current elements form stack
sx.pop();
sy.pop();
// if no path exists
return false;
}
// utility function to check whether path exist or not
bool ispathexist(int srcx, int srcy, int destx, int desty)
{
// to store x co-ordinate
stack sx;
// to store y co-ordinate
stack sy;
return storepath(srcx, srcy, destx, desty, sx, sy);
}
// function to find the path
void printpath(int srcx, int srcy, int destx, int desty)
{
if (!ispathexist(srcx, srcy, destx, desty))
{
// print -1, if path doesn't exist
cout << "-1";
}
}
// driver code
int main()
{
int srcx = 5, srcy = 8;
int destx = 83, desty = 21;
// function call
printpath(srcx, srcy, destx, desty);
}
java 语言(一种计算机语言，尤用于创建网站)
// java program for printing a path from
// given source to destination
import java.util.*;
class gfg{
// function to print the path
static void printexistpath(stack sx,
stack sy, int last)
{
// base condition
if (sx.isempty() || sy.isempty()) {
return;
}
int x = sx.peek();
int y = sy.peek();
// pop stores elements
sx.pop();
sy.pop();
// recursive call for printing stack
// in reverse order
printexistpath(sx, sy, last);
if (sx.size() == last - 1) {
system.out.print("(" x ", " y ")");
}
else {
system.out.print("(" x ", " y ")->");
}
}
// function to store the path into
// the stack, if path exist
static boolean storepath(int srcx, int srcy,
int destx, int desty,
stack sx, stack sy)
{
// base condition
if (srcx > destx || srcy > desty) {
return false;
}
// push current elements
sx.add(srcx);
sy.add(srcy);
// condition to check whether reach to the
// destination or not
if (srcx == destx && srcy == desty) {
printexistpath(sx, sy, sx.size());
return true;
}
// increment 'x' ordinate of source by (2*x y)
// keeping 'y' constant
if (storepath((2 * srcx) srcy, srcy,
destx, desty, sx, sy))
{
return true;
}
// increment 'y' ordinate of source by (2*y x)
// keeping 'x' constant
if (storepath(srcx, (2 * srcy) srcx,
destx, desty, sx, sy))
{
return true;
}
// pop current elements form stack
sx.pop();
sy.pop();
// if no path exists
return false;
}
// utility function to check whether path exist or not
static boolean ispathexist(int srcx, int srcy,
int destx, int desty)
{
// to store x co-ordinate
stack sx = new stack();
// to store y co-ordinate
stack sy = new stack();
return storepath(srcx, srcy, destx,
desty, sx, sy);
}
// function to find the path
static void printpath(int srcx, int srcy,
int destx, int desty)
{
if (!ispathexist(srcx, srcy, destx, desty))
{
// print -1, if path doesn't exist
system.out.print("-1");
}
}
// driver code
public static void main(string[] args)
{
int srcx = 5, srcy = 8;
int destx = 83, desty = 21;
// function call
printpath(srcx, srcy, destx, desty);
}
}
// this code is contributed by 29ajaykumar
python 3
# python3 program for printing a path from
# given source to destination
# function to print the path
def printexistpath( sx, sy, last):
# base condition
if (len(sx) == 0 or len(sy) == 0):
return
x = sx[-1];
y = sy[-1]
# pop stores elements
sx.pop();
sy.pop();
# recursive call for printing stack
# in reverse order
printexistpath(sx, sy, last);
if (len(sx) == last - 1):
print("(" str(x) ", " str(y) ")", end='')
else:
print("(" str(x) ", " str(y) ") -> ", end='')
# function to store the path into
# the stack, if path exist
def storepath(srcx, srcy, destx, desty, sx, sy):
# base condition
if (srcx > destx or srcy > desty):
return false;
# push current elements
sx.append(srcx);
sy.append(srcy);
# condition to check whether reach to the
# destination or not
if (srcx == destx and srcy == desty):
printexistpath(sx, sy, len(sx));
return true;
# increment 'x' ordinate of source by (2*x y)
# keeping 'y' constant
if (storepath((2 * srcx) srcy, srcy, destx, desty, sx, sy)):
return true;
# increment 'y' ordinate of source by (2*y x)
# keeping 'x' constant
if (storepath(srcx, (2 * srcy) srcx, destx, desty, sx, sy)):
return true;
# pop current elements form stack
sx.pop();
sy.pop();
# if no path exists
return false;
# utility function to check whether path exist or not
def ispathexist(srcx, srcy, destx, desty):
# to store x co-ordinate
sx = []
# to store y co-ordinate
sy = []
return storepath(srcx, srcy, destx, desty, sx, sy);
# function to find the path
def printpath(srcx, srcy, destx, desty):
if (not ispathexist(srcx, srcy, destx, desty)):
# print -1, if path doesn't exist
print("-1");
# driver code
if __name__=='__main__':
srcx = 5
srcy = 8;
destx = 83
desty = 21;
# function call
printpath(srcx, srcy, destx, desty);
# this code is contributed by rutvik_56
c
// c# program for printing a path from
// given source to destination
using system;
using system.collections.generic;
class gfg{
// function to print the path
static void printexistpath(stack sx,
stack sy, int last)
{
// base condition
if (sx.count == 0 || sy.count == 0) {
return;
}
int x = sx.peek();
int y = sy.peek();
// pop stores elements
sx.pop();
sy.pop();
// recursive call for printing stack
// in reverse order
printexistpath(sx, sy, last);
if (sx.count == last - 1) {
console.write("(" x ", " y ")");
}
else {
console.write("(" x ", " y ")->");
}
}
// function to store the path into
// the stack, if path exist
static bool storepath(int srcx, int srcy,
int destx, int desty,
stack sx, stack sy)
{
// base condition
if (srcx > destx || srcy > desty) {
return false;
}
// push current elements
sx.push(srcx);
sy.push(srcy);
// condition to check whether reach to the
// destination or not
if (srcx == destx && srcy == desty) {
printexistpath(sx, sy, sx.count);
return true;
}
// increment 'x' ordinate of source by (2*x y)
// keeping 'y' constant
if (storepath((2 * srcx) srcy, srcy,
destx, desty, sx, sy))
{
return true;
}
// increment 'y' ordinate of source by (2*y x)
// keeping 'x' constant
if (storepath(srcx, (2 * srcy) srcx,
destx, desty, sx, sy))
{
return true;
}
// pop current elements form stack
sx.pop();
sy.pop();
// if no path exists
return false;
}
// utility function to check whether path exist or not
static bool ispathexist(int srcx, int srcy,
int destx, int desty)
{
// to store x co-ordinate
stack sx = new stack();
// to store y co-ordinate
stack sy = new stack();
return storepath(srcx, srcy, destx,
desty, sx, sy);
}
// function to find the path
static void printpath(int srcx, int srcy,
int destx, int desty)
{
if (!ispathexist(srcx, srcy, destx, desty))
{
// print -1, if path doesn't exist
console.write("-1");
}
}
// driver code
public static void main(string[] args)
{
int srcx = 5, srcy = 8;
int destx = 83, desty = 21;
// function call
printpath(srcx, srcy, destx, desty);
}
}
// this code is contributed by 29ajaykumar
java 描述语言
output:
(5, 8) -> (5, 21) -> (31, 21) -> (83, 21)
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