实验三

电子老鼠闯迷宫

#include<iostream>
#include<queue>
using namespace std;
char map[15][15];
int sx,sy,ex,ey;
int INF=500;
int d[15][15];
int dx[4]={0,1,0,-1};
int dy[4]={-1,0,1,0};

struct point{
	int x;
	int y;
	point(int x,int y):x(x),y(y){}
};

queue<point> q;

int bfs();

int main(){
	cin>>sx>>sy>>ex>>ey;
	for(int i=1;i<=12;i++){
		for(int j=1;j<=12;j++){
			cin>>map[i][j];
			d[i][j]=INF;
		}
	}
	
	d[sx][sy]=0;
	q.push(point(sx,sy));
	
	cout<<bfs()<<endl;

	return 0;
} 

int bfs(){
	while(!q.empty()){
		point p=q.front();
		q.pop();
		int nx=p.x,ny=p.y;
		if(nx==ex&&ny==ey)break;
		
		for(int i=0;i<4;i++){
			int tx=nx+dx[i],ty=ny+dy[i];
			if(tx>0&&tx<=12
			&&ty>0&&ty<=12
			&&map[tx][ty]!='X'
			&&d[tx][ty]==500){
				q.push(point(tx,ty));
				d[tx][ty]=d[nx][ny]+1;
			}
		}
	}
	return d[ex][ey];
}

跳马

#include<iostream>
#include<queue>
#include<cmath>
#include<cstring>
using namespace std;

int dx[8]={1,2,2,1,-1,-2,-2,-1};
int dy[8]={-2,-1,1,2,2,1,-1,-2};
int derta=3;
int n;

int sx,sy,ex,ey;
int bushu[210][210];//步数很大（0x3f）的话，就是还没有到达过 

struct point{
	int x;
	int y;
	int step;
	point(int x=0,int y=0,int step=0):x(x),y(y),step(step){}

	/*bool operator < (const point &b) {
		int dis=abs(x-ex)+abs(y-ey)+derta*step;
		int disb=abs(b.x-ex)+abs(b.y-ey)+derta*b.step;
		return dis<disb;
	}*/

};

bool operator<(point a,point b){
	int disa=abs(a.x-ex)+abs(a.y-ey)+derta*a.step;
	int disb=abs(b.x-ex)+abs(b.y-ey)+derta*b.step;
	return disa>disb; 	
}

priority_queue<point> q;

int bfs();

int main(){
	cin>>n;
	for(int i=0;i<n;i++){
		memset(bushu,0x3f,sizeof(bushu));
		while(!q.empty())q.pop();
		cin>>sx>>sy>>ex>>ey;
		q.push(point(sx,sy,0));
		bushu[sx][sy]=0; 
		cout<<bfs()<<endl;
	}
	return 0;
}

int bfs(){
	while(!q.empty()){
		point p=q.top();
		q.pop();
		int nx=p.x,ny=p.y,nstep=p.step;
		if(nx==ex&&ny==ey)break;
		for(int i=0;i<8;i++){
			int tx=nx+dx[i],ty=ny+dy[i],tstep=nstep+1;
			if(tx>0&&tx<=200
			&&ty>0&&ty<=200
			&&bushu[tx][ty]==0x3f3f3f3f){
				q.push(point(tx,ty,tstep));
				bushu[tx][ty]=tstep;
			}
		}
	}
	return bushu[ex][ey];
}

独轮车

#include<iostream>
#include<cstring>
#include<queue>
#include<map>
using namespace std;
int dx[4]={0,1,0,-1};
int dy[4]={1,0,-1,0};
char Map[25][25];
int bushu[25][25][10][10];//x坐标，y坐标，颜色，方向 

int sx,sy,sc,sd;char ch_sc,ch_sd;
int ex,ey,ec;char ch_ec; 

struct state{
	int x;
	int y;
	int col;
	int dir;
	state(int x,int y,int col,int dir):x(x),y(y),col(col),dir(dir){}
};

queue<state> q;

int bfs();

int main(){
	map<char,int> mc,md;
	//颜色
	mc['R']=0;mc['Y']=1;mc['B']=2;mc['W']=3;mc['G']=4; 
	//方向 
	md['E']=0;md['S']=1;md['W']=2;md['N']=3;
	
	memset(bushu,0x3f,sizeof(bushu));
	cin>>sx>>sy>>ch_sc>>ch_sd;
	cin>>ex>>ey>>ch_ec;
	for(int i=1;i<=20;i++){
		for(int j=1;j<=20;j++){
			cin>>Map[i][j]; 
		}
	}
	//将字符信息转换为数字信息表示颜色和方向
	sc=mc[ch_sc];
	sd=md[ch_sd];
	ec=mc[ch_ec];
	
	q.push(state(sx,sy,sc,sd));
	bushu[sx][sy][sc][sd]=0;
	cout<<bfs()<<endl;
	return 0;
}

int bfs(){
	int ans=0x3f3f3f3f;//0x3f3f3f3f表示无解 
	while(!q.empty()){
		state s=q.front();
		q.pop();
		int nx=s.x,ny=s.y,nc=s.col,nd=s.dir; 
		if(nx==ex&&ny==ey&&nc==ec){
			ans=bushu[nx][ny][nc][nd];
			break;
		}
		int tx,ty,tc,td;
		//一共有3种变化：向前走，向左转一下，向右转一下 
		//向前走
		tx=nx+dx[nd];
		ty=ny+dy[nd];
		tc=(nc+1)%5;
		td=nd;
		if(tx>=1&&tx<=20
		&&ty>=1&&ty<=20
		&&Map[tx][ty]!='X'
		&&bushu[tx][ty][tc][td]==0x3f3f3f3f){
			q.push(state(tx,ty,tc,td));
			bushu[tx][ty][tc][td]=bushu[nx][ny][nc][nd]+1;
		} 
		//左转
		tx=nx;
		ty=ny;
		tc=nc;
		td=(nd+1)%4;
		if(bushu[tx][ty][tc][td]==0x3f3f3f3f){
			q.push(state(tx,ty,tc,td));
			bushu[tx][ty][tc][td]=bushu[nx][ny][nc][nd]+1;
		} 
		//右转 
		tx=nx;
		ty=ny;
		tc=nc;
		td=(nd+3)%4;
		if(bushu[tx][ty][tc][td]==0x3f3f3f3f){
			q.push(state(tx,ty,tc,td));
			bushu[tx][ty][tc][td]=bushu[nx][ny][nc][nd]+1;
		} 
	}
	return ans;
}

 六数码问题

#include<iostream>
#include<queue>
#include<cstring>
using namespace std;
int visited[7][7][7][7][7][7];//为0则没有访问过 
int s[6],e[6]={1,2,3,4,5,6};

struct state{
	int ar[6];
	state(int a[6]){
		for(int i=0;i<6;i++){
			ar[i]=a[i];
		}
	}
}; 

queue<state> q;

void swap_ar_a(int a[]){
	int t=a[0];
	a[0]=a[3];
	a[3]=a[4];
	a[4]=a[1];
	a[1]=t;
}
void swap_ar_b(int a[]){
	int t=a[1];
	a[1]=a[4];
	a[4]=a[5];
	a[5]=a[2];
	a[2]=t;
} 
bool bfs();

int main(){
	while(cin>>s[0]){
		for(int i=1;i<6;i++)cin>>s[i];
		//先初始化
		memset(visited,0,sizeof(visited));
		while(!q.empty())q.pop(); 
		 
		q.push(state(s));
		visited[s[0]][s[1]][s[2]][s[3]][s[4]][s[5]]=1;
		if(bfs()){
			cout<<"Yes"<<endl;
		}
		else{
			cout<<"No"<<endl;
		}
	}
	return 0;
}

bool bfs(){
	while(!q.empty()){
		state s=q.front();
		q.pop();
		int ns[6];
		int flag=1;//判断是否到达目标状态 
		for(int i=0;i<6;i++){
			ns[i]=s.ar[i];
			if(ns[i]!=e[i])flag=0; 
		}
		if(flag)return true;
		//考察两种变换
		//a变换
		swap_ar_a(ns);
		if(visited[ns[0]][ns[1]][ns[2]][ns[3]][ns[4]][ns[5]]==0){
			q.push(state(ns));
			visited[ns[0]][ns[1]][ns[2]][ns[3]][ns[4]][ns[5]]=1;
		}
		swap_ar_a(ns);swap_ar_a(ns);swap_ar_a(ns);
		//b变换 
		swap_ar_b(ns);
		if(visited[ns[0]][ns[1]][ns[2]][ns[3]][ns[4]][ns[5]]==0){
			q.push(state(ns));
			visited[ns[0]][ns[1]][ns[2]][ns[3]][ns[4]][ns[5]]=1;
		}
		swap_ar_b(ns);swap_ar_b(ns);swap_ar_b(ns);
	}
	return false;
}

加1乘2平方

#include<iostream>
#include<queue>
using namespace std;
int s,e;
int visited[10010];

struct state{
	int num;
	int step;
	state(int num,int step):num(num),step(step){} 
};

queue<state> q;

int bfs();

int main(){
	cin>>s>>e;
	visited[s]=1;
	q.push(state(s,0));
	cout<<bfs()<<endl;
	return 0;
}

int bfs(){
	while(!q.empty()){
		state ns=q.front();
		int n=ns.num,nstep=ns.step;
		q.pop();
		if(n==e)return nstep;
		if(n+1<=e&&visited[n+1]==0){
			visited[n+1]=1;
			q.push(state(n+1,nstep+1));
		}
		if(n*2<=e&&visited[n*2]==0){
			visited[n*2]=1;
			q.push(state(n*2,nstep+1));
		}
		if(n*n<=e&&visited[n*n]==0){
			visited[n*n]=1;
			q.push(state(n*n,nstep+1));
		}
	}
} 

找倍数

#include<iostream>
#include<string> 
#include<queue>
using namespace std;
string ans="1";
int n;

queue<string> q;

bool is_ok(string s,int n);
string bfs();

int main(){
	cin>>n;
	while(n!=0){
		ans="1";
		while(!q.empty())q.pop();
		q.push(ans);
		cout<<bfs()<<endl;
		cin>>n;
	}
	return 0;
}

string bfs(){
	while(!q.empty()){
		string s=q.front();
		//cout<<s<<endl;
		q.pop(); 
		if(!(n%2==0&&s[s.size()-1]=='1')
		&&is_ok(s,n))return s;
		else if(is_ok(s,n))return s; 
		string s1=s+'0';
		q.push(s1);
		string s2=s+'1';
		q.push(s2);
	}
}

bool is_ok(string s,int n){
	int sum=0;
	int size=s.size();
	for(int i=0;i<size;i++){
		sum*=10;
		sum+=s[i]-'0';
		sum%=n;
	}
	if(sum==0)return true;
	else return false; 
}

木乃伊迷宫

#include<iostream>
#include<queue>
using namespace std;
int n;
int dx,dy,dtype;
int msx,msy;//木乃伊
int rsx,rsy;//人 
int ex,ey;//出口
int visited[6][6][6][6]; 
int Dx[4]={0,1,0,-1};
int Dy[4]={-1,0,1,0};
 
struct point{
	int door[4];
	point(int a=0,int b=0,int c=0,int d=0){
		door[0]=a;door[1]=b;door[2]=c;door[3]=d;
	}
};

point map[6][6];

struct state{
	int mx,my;
	int rx,ry;
	state(int a,int b,int c,int d):mx(a),my(b),rx(c),ry(d){}
};

queue<state> q;

bool bfs();

int main(){
	cin>>n;
	for(int i=0;i<n;i++){
		cin>>dx>>dy>>dtype;
		if(dtype==0){
			map[dx][dy].door[1]=1;
			if(dx+1<6)map[dx+1][dy].door[3]=1;
		}
		else if(dtype==1){
			map[dx][dy].door[2]=1;
			if(dx+1<6)map[dx][dy+1].door[0]=1;
		}
	}
	cin>>msx>>msy>>rsx>>rsy>>ex>>ey;
	q.push(state(msx,msy,rsx,rsy));
	visited[msx][msy][rsx][rsy]=1;
	if(bfs()){
		cout<<"Yes"<<endl;
	}
	else{
		cout<<"No"<<endl;
	}
	return 0;
}

bool bfs(){
	while(!q.empty()){
		state s=q.front();
		q.pop();
		int mnx=s.mx,mny=s.my,rnx=s.rx,rny=s.ry;
		//cout<<mnx<<' '<<mny<<' '<<rnx<<' '<<rny<<endl;
		if(rnx==ex&&rny==ey)return true;
		//人的一步
		for(int i=0;i<4;i++){
			if(mny==rny&&mnx==rnx)continue; 
			int rtx=rnx+Dx[i],rty=rny+Dy[i];
			int flag=1;//表示该状态可行 
			if(!(rtx>=0&&rtx<6
			&&rty>=0&&rty<6
			&&map[rnx][rny].door[i]==0))continue;
			if(mny==rty&&mnx==rtx)continue; 
			//木乃伊的两步
			mnx=s.mx,mny=s.my;	
			for(int j=0;j<2;j++){
				if(rty==mny){//同一列 
					if(rtx<mnx)//人在上面
					{
					if(map[mnx][mny].door[3]==0)mnx--;	
					 } 
					else if(rtx>mnx)//人在下面
					{
					if(map[mnx][mny].door[1]==0)mnx++;
					 } 
				}
				else if(rty<mny){//木乃伊需要往左移动 
					if(map[mnx][mny].door[0]==0)mny--;
				} 
				else if(rty<mny){//木乃伊需要向右移动 
					if(map[mnx][mny].door[2]==0)mny++;
				}
				//考察这一步走后是否木乃伊会碰到人 
				if(mny==rty&&mnx==rtx){
					flag=0;
					break;
				}	
			}
			if(flag==0)continue;//不可行的状态，就要考察其他状态 
			//木乃伊走完，考察状态
			if(visited[mnx][mny][rtx][rty]==0){
				visited[mnx][mny][rtx][rty]=1;
				q.push(state(mnx,mny,rtx,rty));
			} 
		}		
	}
	return false;
}

polygon

#include<iostream>
#include<cmath>
#include<iomanip>
using namespace std;
int n,m;
double nxy[1010];
double ans=0; 
int main(){
	cin>>n>>m;
	//一共有n+m个坐标
	//求出n个点每一个的坐标
	for(int i=1;i<=n;i++){
		nxy[i]=(n+m)/double(n)*i;
	}
	//统计所有坐标到最近位置的距离
	for(int i=1;i<=n;i++){
		ans+=min(  ceil(nxy[i])-nxy[i] , nxy[i]-floor(nxy[i])  );
	}
	//换算成标准距离，输出 
	cout<<fixed<<setprecision(4)<<ans*10000.0/(n+m)<<endl;
	return 0;
}

推箱子

#include<iostream>
#include<queue>
#include<cmath>
#include<string>
using namespace std;
int map[20][20];
string s[11];
int ex,ey;
int rsx,rsy;
int bsx,bsy;
int visited[11][11][11][11];

struct state{
	int rx,ry;
	int bx,by;
	int step;
	state(int a,int b,int c,int d,int e):rx(a),ry(b),bx(c),by(d),step(e){}
};

queue<state> q;

int bfs();

int main(){
	for(int i=0;i<10;i++)cin>>s[i];
	for(int i=1;i<=10;i++){
		for(int j=1;j<=10;j++){
			map[i][j]=(s[i-1][j-1]-'0');
			if(map[i][j]==2){
				bsx=i;
				bsy=j;
			}
			else if(map[i][j]==3){
				ex=i;
				ey=j;
			}
			else if(map[i][j]==4){
				rsx=i;
				rsy=j;
			}
		}
	}
	q.push(state(rsx,rsy,bsx,bsy,0));
	visited[rsx][rsy][bsx][bsy]=1;
	cout<<bfs()<<endl;
	return 0;
}

int bfs(){
	while(!q.empty()){
		state s=q.front();
		int rnx=s.rx,rny=s.ry,bnx=s.bx,bny=s.by,nstep=s.step;
		//cout<<rnx<<' '<<rny<<' '<<bnx<<' '<<bny<<' '<<nstep<<endl; 
		q.pop();
		if(bnx==ex&&bny==ey)return nstep;
		//一下枚举所有移动
		//人和箱子靠在一起且可以往另一边移动 
		if(rnx==bnx&&rny+1==bny&&rny+2<=10&&visited[rnx][rny+1][rnx][rny+2]==0&&map[rnx][rny+2]!=1){
			q.push(state(rnx,rny+1,rnx,rny+2,nstep+1));
			visited[rnx][rny+1][rnx][rny+2]=1;
		}
		if(rnx==bnx&&rny-1==bny&&rny-2>=1&&visited[rnx][rny-1][rnx][rny-2]==0&&map[rnx][rny-2]!=1){
			q.push(state(rnx,rny-1,rnx,rny-2,nstep+1));
			visited[rnx][rny-1][rnx][rny-2]=1;
		}
		if(rny==bny&&rnx+1==bnx&&rnx+2<=10&&visited[rnx+1][rny][rnx+2][rny]==0&&map[rnx+2][rny]!=1){
			q.push(state(rnx+1,rny,rnx+2,rny,nstep+1));
			visited[rnx+1][rny][rnx+2][rny]=1;
		}
		if(rny==bny&&rnx-1==bnx&&rnx-2>=1&&visited[rnx-1][rny][rnx-2][rny]==0&&map[rnx-2][rny]!=1){
			q.push(state(rnx-1,rny,rnx-2,rny,nstep+1));
			visited[rnx-1][rny][rnx-2][rny]=1;
		} 
		//接下来就是人不推箱子，只是走一走的过程
		//按照左下右上顺序
		//注意人不能穿过箱子 
		if(rny-1>=1&&visited[rnx][rny-1][bnx][bny]==0&&map[rnx][rny-1]!=1&&!(rnx-bnx==0&&rny-1-bny==0)){
			q.push(state(rnx,rny-1,bnx,bny,nstep+1));
			visited[rnx][rny-1][bnx][bny]=1;
		}
		if(rnx+1<=10&&visited[rnx+1][rny][bnx][bny]==0&&map[rnx+1][rny]!=1&&!(rnx+1-bnx==0&&rny-bny==0)){
			q.push(state(rnx+1,rny,bnx,bny,nstep+1));
			visited[rnx+1][rny][bnx][bny]=1;
		}
		if(rny+1<=10&&visited[rnx][rny+1][bnx][bny]==0&&map[rnx][rny+1]!=1&&!(rnx-bnx==0&&rny+1-bny==0)){
			q.push(state(rnx,rny+1,bnx,bny,nstep+1));
			visited[rnx][rny+1][bnx][bny]=1;
		}
		if(rnx-1>=1&&visited[rnx-1][rny][bnx][bny]==0&&map[rnx-1][rny]!=1&&!(rnx-1-bnx==0&&rny-bny==0)){
			q.push(state(rnx-1,rny,bnx,bny,nstep+1));
			visited[rnx-1][rny][bnx][bny]=1;
		}
	}
	return -1;
}

特殊的二阶魔方

#include<iostream>
#include<map>
#include<queue>
using namespace std;
int t[2];
char s[6][4];
int v[16][16][16][16][16][16];
struct state{
	char f[6][4];
	int num[6];
	int step;
	state(char ch[6][4],int s){
		for(int i=0;i<6;i++){
			num[i]=0;
			for(int j=0;j<4;j++){
				f[i][j]=ch[i][j];
				num[i]*=2;
				num[i]+=ch[i][j]-'0';
			}
		}
		step=s;
	}
};
queue<state> q;
void visited(state s){
	int a=s.num[0],b=s.num[1],c=s.num[2],d=s.num[3],e=s.num[4],f=s.num[5];
	v[a][b][c][d][e][f]=1;
}
bool isnt_visited(state s){
	int a=s.num[0],b=s.num[1],c=s.num[2],d=s.num[3],e=s.num[4],f=s.num[5];
	if(v[a][b][c][d][e][f]==0)return true;
	else return false;
}
void fengzhuang(char c[6][4],int step){
	state t(c,step+1);
	if(isnt_visited(t)){
		visited(t);
		q.push(t);
	}
}
void sw(char c[6][4],int ai,int aj,int bi,int bj,int ci,int cj,int di,int dj){
	int t=c[ai][aj];
	c[ai][aj]=c[di][dj];
	c[di][dj]=c[ci][cj];
	c[ci][cj]=c[bi][bj];
	c[bi][bj]=t;
} 
void b1(char c[6][4]);
void b2(char c[6][4]);
void b3(char c[6][4]);
int bfs();


int main(){
	for(int i=0;i<6;i++){
		for(int j=0;j<2;j++){
			cin>>t[j];
		}
		s[i][0]=t[0]/10+'0';
		s[i][1]=t[0]%10+'0';
		s[i][2]=t[1]/10+'0';
		s[i][3]=t[1]%10+'0';
	}
	/*for(int i=0;i<6;i++){
		for(int j=0;j<4;j++){
			cout<<s[i][j];
		}
		cout<<endl;
	}*/
	state st(s,0);
	visited(st);
	q.push(st);
	cout<<bfs()<<endl;
	return 0;
}



void b1(char c[6][4]){
	sw(c,2,2,5,3,3,3,4,2);
	sw(c,2,3,5,2,3,2,4,3);
	sw(c,1,2,1,0,1,1,1,3);
}
void b2(char c[6][4]){
	sw(c,1,3,4,1,0,1,5,3);
	sw(c,1,1,4,3,0,3,5,1);
	sw(c,3,0,3,2,3,3,3,1);
}
void b3(char c[6][4]){
	sw(c,2,0,0,1,3,2,1,0);
	sw(c,2,2,0,0,3,0,1,1);
	sw(c,5,2,5,0,5,1,5,3);
}
int bfs(){
	while(!q.empty()){
		state n=q.front();
		q.pop();
		char fn[6][4];
		int nstep=n.step;
		for(int i=0;i<6;i++){
			int flag=1;
			for(int j=0;j<4;j++){
				fn[i][j]=n.f[i][j];
				if(fn[i][j]!='0')flag=0;
			}
			if(flag)return nstep;
		}
		/*for(int i=0;i<6;i++){
			for(int j=0;j<4;j++){
				cout<<fn[i][j];
			}
			cout<<endl;
		}*/
		//1：得到变换之后的字符串数组
		//2：得到状态
		//3：判断状态是否重复
		//4：没有重复则加入队列
		//5：2~4都封装到fengzhuang函数里面进行 
		
		//1
		b1(fn);
		fengzhuang(fn,nstep);
		//2
		b1(fn);b1(fn);
		fengzhuang(fn,nstep);
		b1(fn);
		//3
		b2(fn);
		fengzhuang(fn,nstep);
		//4
		b2(fn);b2(fn);
		fengzhuang(fn,nstep);
		b2(fn);
		//5
		b3(fn);
		fengzhuang(fn,nstep);
		//6
		b3(fn);b3(fn);
		fengzhuang(fn,nstep);
		b3(fn);
	}
	return -1;
}