逻辑题：假设河的同一侧有警察、强盗……

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假设河的同一侧有警察、强盗、路人A、路人A的两个孩子（A1，A2）、路人B、路人B的两个孩子（B1、B2），他们想要安全过河，限制条件如下：
1、假设警察不在，强盗会杀死所有的跟他在一起的人
2、假设路人A离开，路人B和路人A的孩子在一起，路人A的孩子会死掉
3、假设路人B离开，路人A和路人B的孩子在一起，路人B的孩子会死掉
4、河面上有一条船，但一次最多可以容纳两人过河
5、只有警察、路人A、路人B会划船
请给出你的过河方案。

#include <bits/stdc++.h>using namespace std;struct state{int l[8];int r[8];int boat;state(){for (int i=0;i<8;i++)l[i]=1;memset(r,0,sizeof(r));boat=0;}};bool ok(state s){bool flag=true;for (int i=0;i<8;i++)if (!s.r[i]){flag=false;break;}return flag;}bool valid(state s){bool flag=true;//left validif (!s.l[0] && s.l[1] && (s.l[2]||s.l[3]||s.l[4]||s.l[5]||s.l[6]||s.l[7]))flag=false;if (s.l[2] && !s.l[5] && s.l[6] && s.l[7])flag=false;if (s.l[5] && !s.l[2] && s.l[3] && s.l[4])flag=false;//right validif (!s.r[0] && s.r[1] && (s.r[2]||s.r[3]||s.r[4]||s.r[5]||s.r[6]||s.r[7]))flag=false;if (s.r[2] && !s.r[5] && s.r[6] && s.r[7])flag=false;if (s.r[5] && !s.r[2] && s.r[3] && s.r[4])flag=false;return flag;}const char str[][100]={"警察","强盗","A","A1","A2","B","B1","B2"};string path[100];bool found=false;map<string, bool> M;string toString(state s){string temp;for (int i=0;i<8;i++){temp+=s.l[i]+'0';temp+=s.r[i]+'0';}temp+=s.boat+'0';//cout << temp << endl;return temp;}void dfs(state s,int depth){if (M[toString(s)]) return;if (found) return;if (!valid(s)) return;if (ok(s)){for (int i=0;i<depth;i++){if (i%2==0)cout << "left --> right : ";elsecout << "left <-- right : ";cout << path[i] << endl;}cout << "DONE" << endl;found=true;return;}M[toString(s)]=true;if (!s.boat)//left{if (s.l[0]){state newState=s;newState.boat=s.boat^1;newState.l[0]=0;newState.r[0]=1;path[depth]=str[0];dfs(newState,depth+1);for (int i=0;i<8;i++)if (i!=0 && s.l[i]){state newState=s;newState.boat=s.boat^1;newState.l[i]=newState.l[0]=0;newState.r[i]=newState.r[0]=1;path[depth]=str[0];path[depth]+=" ";path[depth]+=str[i];dfs(newState,depth+1);}}if (s.l[2]){state newState=s;newState.boat=s.boat^1;newState.l[2]=0;newState.r[2]=1;path[depth]=str[2];dfs(newState,depth+1);for (int i=0;i<8;i++)if (i!=2 && s.l[i]){state newState=s;newState.boat=s.boat^1;newState.l[i]=newState.l[2]=0;newState.r[i]=newState.r[2]=1;path[depth]=str[2];path[depth]+=" ";path[depth]+=str[i];dfs(newState,depth+1);}}if (s.l[5]){state newState=s;newState.boat=s.boat^1;newState.l[5]=0;newState.r[5]=1;path[depth]=str[5];dfs(newState,depth+1);for (int i=0;i<8;i++)if (i!=5 && s.l[i]){state newState=s;newState.boat=s.boat^1;newState.l[i]=newState.l[5]=0;newState.r[i]=newState.r[5]=1;path[depth]=str[5];path[depth]+=" ";path[depth]+=str[i];dfs(newState,depth+1);}}}else//right{if (s.r[0]){state newState=s;newState.boat=s.boat^1;newState.r[0]=0;newState.l[0]=1;path[depth]=str[0];dfs(newState,depth+1);for (int i=0;i<8;i++)if (i!=0 && s.r[i]){state newState=s;newState.boat=s.boat^1;newState.r[i]=newState.r[0]=0;newState.l[i]=newState.l[0]=1;path[depth]=str[0];path[depth]+=" ";path[depth]+=str[i];dfs(newState,depth+1);}}if (s.r[2]){state newState=s;newState.boat=s.boat^1;newState.r[2]=0;newState.l[2]=1;path[depth]=str[2];dfs(newState,depth+1);for (int i=0;i<8;i++)if (i!=2 && s.r[i]){state newState=s;newState.boat=s.boat^1;newState.r[i]=newState.r[2]=0;newState.l[i]=newState.l[2]=1;path[depth]=str[2];path[depth]+=" ";path[depth]+=str[i];dfs(newState,depth+1);}}if (s.r[5]){state newState=s;newState.boat=s.boat^1;newState.r[5]=0;newState.l[5]=1;path[depth]=str[5];dfs(newState,depth+1);for (int i=0;i<8;i++)if (i!=5 && s.r[i]){state newState=s;newState.boat=s.boat^1;newState.r[i]=newState.r[5]=0;newState.l[i]=newState.l[5]=1;path[depth]=str[5];path[depth]+=" ";path[depth]+=str[i];dfs(newState,depth+1);}}}}int main(){state init;dfs(init,0);}

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