B – Dungeon MasterTime Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64uSubmit Status Practice POJ 2251DescriptionYou are trapped in a 3D dungeon and need to find the quickest way out! The dungeon is composed of unit cubes which may or may not be filled with rock. It takes one minute to move one unit north, south, east, west, up or down. You cannot move diagonally and the maze is surrounded by solid rock on all sides.Is an escape possible? If yes, how long will it take?InputThe input consists of a number of dungeons. Each dungeon description starts with a line containing three integers L, R and C (all limited to 30 in size).L is the number of levels making up the dungeon.R and C are the number of rows and columns making up the plan of each level.Then there will follow L blocks of R lines each containing C characters. Each character describes one cell of the dungeon. A cell full of rock is indicated by a ‘#’ and empty cells are represented by a ‘.’. Your starting position is indicated by ‘S’ and the exit by the letter ‘E’. There’s a single blank line after each level. Input is terminated by three zeroes for L, R and C.OutputEach maze generates one line of output. If it is possible to reach the exit, print a line of the formEscaped in x minute(s). where x is replaced by the shortest time it takes to escape.If it is not possible to escape, print the lineTrapped! Sample Input3 4 5S…..###..##..###.#############.####…###########.#######E1 3 3S###E####0 0 0Sample OutputEscaped in 11 minute(s).
Trapped!
//类比二维的图的搜索即可，，但是注意三维图的存储细节
#include<iostream>#include<cstring>#include<string>#include<cmath>#include<map>#include<queue>#include<cstdio>#include<vector>#include<algorithm>using namespace std;const int maxn=100;const int inf=210000;typedef long long ll;int x,y,z;char a[maxn][maxn][maxn];bool vis[maxn][maxn][maxn];int d[maxn][maxn][maxn];struct node{int x,y,z;node(int a,int b,int c){x=a;y=b;z=c;}node(){}};node end;int xx[]={-1,0,0,1,0,0},yy[]={0,-1,0,0,1,0},zz[]={0,0,-1,0,0,1};void bfs(int sx,int sy,int sz){node tmp=node(sx,sy,sz);queue<node> q;d[sx][sy][sz]=0;q.push(tmp);vis[sx][sy][sz]=true;while(!q.empty()){tmp=q.front();q.pop();if(tmp.x==end.x&&tmp.y==end.y&&tmp.z==end.z){return;}for(int i=0;i<6;i++){int dx=tmp.x+xx[i];int dy=tmp.y+yy[i];int dz=tmp.z+zz[i];if(dx<x&&dx>=0&&dy<y&&dy>=0&&dz<z&&dz>=0&&!vis[dx][dy][dz]&&a[dx][dy][dz]!='#'){vis[dx][dy][dz]=true;q.push(node(dx,dy,dz));d[dx][dy][dz]=d[tmp.x][tmp.y][tmp.z]+1;}}}}int main(){int m,i,j,t,k;//freopen("in.txt","r",stdin);while(cin>>z>>x>>y&&(x||y||z)){int sx,sy,sz;for(i=0;i<z;i++)for(j=0;j<x;j++)for(k=0;k<y;k++){cin>>a[j][k][i];if(a[j][k][i]=='S'){sx=j;sy=k;sz=i;}else if(a[j][k][i]=='E'){end.x=j;end.y=k;end.z=i;}}//cout<<sx<<' '<<sy<<' '<<sz<<endl;// cout<<end.x<<' '<<end.y<<' '<<end.z<<endl;memset(vis,0,sizeof(vis));memset(d,0,sizeof(d));bfs(sx,sy,sz);if(d[end.x][end.y][end.z]==0)printf("Trapped!\n");elseprintf("Escaped in %d minute(s).\n",d[end.x][end.y][end.z]);}return 0;}
人生就像一杯没有加糖的咖啡，喝起来是苦涩的，
