牧师和野人过河的问题，请高手指点我的程序错误（cpp&python）

[hugsnow]

CPP 代码

#include <stdio.h>
#include <vector>
#include <algorithm>
#include <iostream>

using namespace std;
bool ok(int a,int b);
bool guoqu(int a,int b,vector < int > c);
bool guolai(int a,int b,vector < int > c);
const int ren=3;
const int sheep=3;
int main(int argc, char **argv)
{
	vector < int > v;
	v.push_back(33);
	guoqu(3,3,v);
	system("pause");
	return 0;
}
bool guoqu(int a,int b,vector < int > c){
	//if(a==0&&b==0)
	//	return true;
	for(int i=0;i<=a;i++)
		for(int j=0;j<=b;j++)
			if(i+j>0&&i+j<=sheep){
				int a1,b1,a2,b2,d;
				a1=a-i;
				b1=b-j;
				a2=ren-a1;
				b2=ren-b1;
				d=a1*10+b1;
				if(find(c.begin(),c.end(),d)==c.end()&&ok(a1,b1)){
					c.push_back(d);
					if(guolai(a2,b2,c)){
						cout<<c.size()-1<<".运"<<i<<"牧师和"<<j<<"野人过去,"
							<<"这边"<<a1<<b1<<",那边"<<a2<<b2<<endl;
						return true;
					}
				}
			}
	return false;
}
bool guolai(int a,int b,vector < int > c){
	if(a==3&&b==3)
		return true;
	for(int i=0;i<=a;i++)
		for(int j=0;j<=b;j++)
			if(i+j>0&&i+j<=sheep){
				int a1,b1,a2,b2,a3,b3,d;
				a1=a-i;
				b1=b-j;
				a2=ren-a1;
				b2=ren-b1;
				d=a2*10+b2;
				if(find(c.begin(),c.end(),d)==c.end()&&ok(a2,b2)){
					c.push_back(d);
					if(guoqu(a2,b2,c)){
						cout<<c.size()-1<<".运"<<i<<"牧师和"<<j<<"野人过来,"
							<<"这边"<<a2<<b2<<",那边"<<a1<<b1<<endl;
						return true;
					}
				}
			}
	return false;
}

bool ok(int a,int b){
	return a==3||a==0||a==b;
}

Python 代码:

ren=3;
chuan=2;
def ok(a,b):
  (a==3) or (a==0) or (a==b)
  
def guoqu(a,b,c):
  for ms in range(a+1):
    for yr in range(b+1):
      if (ms+yr>0) and (ms+yr<=chuan):
        zbms=a-ms
        zbyr=b-yr
        nbms=ren-zbms
        nbyr=ren-zbyr
        shu=zbms*10+zbyr
        if (shu not in c) and (ok(zbms,zbyr)):
          if guolai(nbms,nbyr,c+[shu]):
            print "%d and %d,this:%d%d,that:%d%d"%(ms,yr,zbms,zbyr,nbms,nbyr)
            return True
  return False
  
def guolai(a,b,c):
  if (a==3) and (b==3):
    return True
  for ms in range(a+1):
    for yr in range(b+1):
      if (ms+yr>0) and (ms+yr<=chuan):
        nbms=a-ms
        nbyr=b-yr
        zbms=ren-nbms
        zbyr=ren-nbyr
        shu=zbms*10+zbyr
        if (shu not in c) and (ok(zbms,zbyr)):
          if guoqu(zbms,zbyr,c+[shu]):
            print "%d and %d,this:%d%d,that:%d%d"%(ms,yr,zbms,zbyr,nbms,nbyr)
            return True
  return False
  
guoqu(0,2,[33])

[hugsnow]

修正：const int sheep=2

[tailzhou]

你的判断状态重复的方式是错的；

除了判断这边岸上的数目外，还需要判断船在哪一边；

比如：这边岸上有3个牧师，一个野人1;船在对岸与船在这边岸上是两个不同的状态；

可以用个3位数来表示状态，最后一位表示船的状态，初始状态为331,1表示船在这边岸上,0表示在对岸；

bool guoqu(int a,int b,vector < int > c){
...
d=a1*10+b1;
==》
d=a1*100+b1*10+0;

bool guolai(int a,int b,vector < int > c){
...
d=a2*10+b2;
==>
d=a2*100+b2*10+1;

[hugsnow]

谢谢，果然如此，给分

[hugsnow]

如果是这样，那么没有必要区分过去还是过来，直接一个函数就可以解决了，本来我就写的一个函数，就是因为上面的错误不成功才分成两个函数。