蓝桥杯-地牢大师
作者:互联网
链接:https://www.acwing.com/problem/content/1098/
思路:1:三维bfs:可以分为当前层数内的二维内偏移和层数的一维偏移
代码:
#include<iostream>
#include<queue>
#include<string>
#include<cstring>
using namespace std;
int l,r,c;
char s[105][105][105];
int b[105][105][105];
int dx[]={0,-1,0,1};
int dy[]={-1,0,1,0};
int dz[]={-1,1};
struct node{
int x,y,z;
node(int _z,int _x,int _y):x(_x),y(_y),z(_z){}
};
void bfs(int z,int x,int y){
int ta,tb,tc;
queue<node>q;
q.push(node(z,x,y));
memset(b,-1,sizeof(b));
b[z][x][y]=1;
while(q.size()){
x=q.front().x;
y=q.front().y;
z=q.front().z;
q.pop();
for(int i=0;i<4;i++){
ta=z;
tb=x+dx[i];
tc=y+dy[i];
if(ta<1||tb<1||tc<1||ta>l||tb>r||tc>c||s[ta][tb][tc]=='#')continue;
if(b[ta][tb][tc]!=-1)continue;
// cout<<ta<<" "<<tb<<" "<<tc<<endl;
b[ta][tb][tc]=b[z][x][y]+1;
q.push(node(ta,tb,tc));
if(s[ta][tb][tc]=='E')return;
}
for(int i=0;i<2;i++){
ta=z+dz[i];
tb=x;
tc=y;
if(ta<1||tb<1||tc<1||ta>l||tb>r||tc>c||s[ta][tb][tc]=='#')continue;
if(b[ta][tb][tc]!=-1)continue;
//cout<<ta<<" "<<tb<<" "<<tc<<endl;
b[ta][tb][tc]=b[z][x][y]+1;
q.push(node(ta,tb,tc));
if(s[ta][tb][tc]=='E')return;
}
}
}
int main (){
int x,y,z;
while(cin>>l>>r>>c&&(l+r+c)){
for(int t=1;t<=l;t++){
for(int i=1;i<=r;i++)
cin>>(s[t][i]+1);
}
for(int t=1;t<=l;t++)
for(int i=1;i<=r;i++)
for(int j=1;j<=c;j++)
if(s[t][i][j]=='S')bfs(t,i,j);
else if(s[t][i][j]=='E'){
x=t;
y=i;
z=j;
}
if(b[x][y][z]==-1)cout<<"Trapped!"<<endl;
else cout<<"Escaped in "<<b[x][y][z]-1<<" minute(s)."<<endl;
}
return 0;
}
标签:地牢,include,int,ta,蓝桥,105,tb,tc,大师 来源: https://www.cnblogs.com/abestxun/p/14661402.html