Description of the problem:
[Problem of missionaries and cannibals]
There are 3 missionaries, 1 boat and 3 cannibals on the left bank of the river. The missionaries want to use this boat to transport all members to the right bank of the river, but Subject to the following conditions:
(1) The boat can only hold a maximum of 2 passengers at a time (missionaries and cannibals will both row);
(2) On any shore, if the number of cannibals exceeds the number of missionaries, the missionaries will be Cannibals eat.
(3) Assume that cannibals will obey any kind of river-crossing arrangement.
C++ version:
#include<iostream>
#include<queue>
using namespace std;
int m,c,k;//m个传教士，n个食人者，船上可以有k个人
int xx[]={
0,0,1,1,2};//每次可以上船的人员数量
int yy[]={
1,2,0,1,0};
struct zt{
int lm;
int lc;
int rm;
int rc;
int ship;
int top;
};
queue<zt>open;
struct zt temp[1000];
int index=0;
int Isdangerous(zt temp)
{
if((temp.lm>=temp.lc&&temp.rm>=temp.rc&&temp.lm>=0&&temp.lc>=0&&temp.rm>=0&&temp.rc>=0)||
(temp.lm>=0&&temp.lc>=0&&temp.rm>=0&&temp.rc>=0&&temp.lm==0&&temp.rm>=temp.rc)||
(temp.lm>=0&&temp.lc>=0&&temp.rm>=0&&temp.rc>=0&&temp.lm>=temp.lc&&temp.rm==0))
{
return 1;
}
return 0;
}
int pd(zt p,int top)
{
for(int i=0;i<top;i++)
{
if(temp[i].lm==p.lm&&temp[i].lc==p.lc&&temp[i].rm==p.rm&&temp[i].rc==p.rc&&temp[i].ship==p.ship)
{
return 0;
}
}
return 1;
}
void bfS()
{
while(!open.empty())
{
zt temp1=open.front();
int x1=temp1.lm;
int y1=temp1.lc;
int x2=temp1.rm;
int y2=temp1.rc;
for(int i=0;i<5;i++)
{
if(temp1.ship==1){
zt temp2;
temp2.lm=x1-xx[i];
temp2.lc=y1-yy[i];
temp2.rm=x2+xx[i];
temp2.rc=y2+yy[i];
temp2.ship=-temp1.ship;
temp2.top=temp1.top+1;
if(temp2.lm==0&&temp2.lc==0&&temp2.rm==3&&temp2.rc==3)
{
goto loop;
}
if(Isdangerous(temp2)==1&&pd(temp2,index)==1)
{
temp[index++]=temp2;
open.push(temp2);
}
}
if(temp1.ship==-1){
zt temp2;
temp2.lm=x1+xx[i];
temp2.lc=y1+yy[i];
temp2.rm=x2-xx[i];
temp2.rc=y2-yy[i];
temp2.ship=-temp1.ship;
temp2.top=temp1.top+1;
if(Isdangerous(temp2)==1&&pd(temp2,index)==1)
{
temp[index++]=temp2;
open.push(temp2);
}
}
}
open.pop();
}
loop:;
}
int main()
{
zt p;
p.lm=3;
p.lc=3;
p.rm=0;
p.rc=0;
p.ship=1;
p.top=0;
open.push(p);
temp[index++]=p;
bfS();
cout<<"初始状态:"<<3<<' '<<3<<' '<<0<<' '<<0<<endl;
for(int i=3;i<index-1;i++)
{
if(temp[i].ship==1)
{
cout<<"B----->A:";
cout<<temp[i].lm<<' '<<temp[i].lc<<' '<<temp[i].rm<<' '<<temp[i].rc<<endl;
}
if(temp[i].ship==-1)
{
cout<<"A----->B:";
cout<<temp[i].lm<<' '<<temp[i].lc<<' '<<temp[i].rm<<' '<<temp[i].rc<<endl;
}
}
cout<<"目标状态:"<<0<<' '<<0<<' '<<3<<' '<<3<<endl;
return 0;
}
python version:
from queue import Queue
class zt:
def __init__(self,x1,y1,x2,y2,ship,top):
self.lm=x1
self.lc=y1
self.rm=x2
self.rc=y2
self.ship=ship
self.top=top
temp = Queue(maxsize=0)
xx=[0,0,1,1,2]
yy=[1,2,0,1,0]
list=[]#对象list
index=0#对象指针
def Isdangerous(temp):
if (( temp.lm >= temp.lc and temp.rm >= temp.rc and temp.lm >= 0 and temp.lc >= 0 and temp.rm >= 0 and temp.rc >= 0)
or (temp.lm >= 0 and temp.lc >= 0and temp.rm >= 0 and temp.rc >= 0 and temp.lm == 0 and temp.rm >= temp.rc)
or (temp.lm >= 0 and temp.lc >= 0 and temp.rm >= 0 and temp.rc >= 0 and temp.lm >= temp.lc and temp.rm == 0)):
return 1
else:
return 0
def pd(p,index):
for i in range(0,index):
if(list[i].lm==p.lm and list[i].lc==p.lc and list[i].rm==p.rm and list[i].rc==p.rc and list[i].ship==p.ship):
return 0
return 1
def fin(temp):
if temp.lm==0 and temp.lc==0 and temp.rm==3 and temp.rc==3:
return 1
return 0
def bfs():
global index
while temp.empty()==0:
temp1=temp.get()
x1=temp1.lm
y1=temp1.lc
x2=temp1.rm
y2=temp1.rc
for i in range(0,5):
x=xx[i]
y=yy[i]
if temp1.ship==1:
temp2=zt(x1-x,y1-y,x2+x,y2+y,-1,temp1.top+1)
if fin(temp2)==1:
return
if Isdangerous(temp2)==1and pd(temp2,index)==1:
temp.put(temp2)
list.append(temp2)
index+=1
else :
temp2 = zt(x1 + x, y1 + y, x2 - x, y2 - y, 1, temp1.top + 1)
if Isdangerous(temp2) == 1 and pd(temp2, index) == 1:
temp.put(temp2)
list.append(temp2)
index += 1
start=zt(3,3,0,0,1,0)
temp.put(start)
list.append(start)
index+=1
bfs()
print('初始状态',3,3,0,0)
for item in range(3,index-1):
if list[item].ship==-1:
if item==3:
print(" A----->B:", end='')
else:
print("A----->B:",end='')
print(list[item].lm,list[item].lc,list[item].rm,list[item].rc,'\n',end=' ')
if list[item].ship==1:
print("B----->A:",end='')
print(list[item].lm,list[item].lc,list[item].rm,list[item].rc,'\n',end=' ')
print("目标状态:",0,0,3,3)