Asteroids!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3559 Accepted Submission(s): 2360
You want to get home.
There are asteroids.
You don't want to hit them.
A single data set has 5 components:
Start line - A single line, "START N", where 1 <= N <= 10.
Slice list - A series of N slices. Each slice is an N x N matrix representing a horizontal slice through the asteroid field. Each position in the matrix will be one of two values:
'O' - (the letter "oh") Empty space
'X' - (upper-case) Asteroid present
Starting Position - A single line, "A B C", denoting the <A,B,C> coordinates of your craft's starting position. The coordinate values will be integers separated by individual spaces.
Target Position - A single line, "D E F", denoting the <D,E,F> coordinates of your target's position. The coordinate values will be integers separated by individual spaces.
End line - A single line, "END"
The origin of the coordinate system is <0,0,0>. Therefore, each component of each coordinate vector will be an integer between 0 and N-1, inclusive.
The first coordinate in a set indicates the column. Left column = 0.
The second coordinate in a set indicates the row. Top row = 0.
The third coordinate in a set indicates the slice. First slice = 0.
Both the Starting Position and the Target Position will be in empty space.
A single output set consists of a single line. If a route exists, the line will be in the format "X Y", where X is the same as N from the corresponding input data set and Y is the least number of moves necessary to get your ship from the starting position to
the target position. If there is no route from the starting position to the target position, the line will be "NO ROUTE" instead.
A move can only be in one of the six basic directions: up, down, left, right, forward, back. Phrased more precisely, a move will either increment or decrement a single component of your current position vector by 1.
START 1 O 0 0 0 0 0 0 END START 3 XXX XXX XXX OOO OOO OOO XXX XXX XXX 0 0 1 2 2 1 END START 5 OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO XXXXX XXXXX XXXXX XXXXX XXXXX OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO 0 0 0 4 4 4 END
1 0 3 4 NO ROUTE
题解:三维的迷宫问题,没什么难度,需要注意的是维度为map[z][x][y]
*/
#include<cstdio> #include<cstring> #include<queue> #define inf 0x3f3f3f3f using namespace std; int n,vis[12][12][12],ex,ey,ez; int dir[6][3]={{0,0,1},{0,0,-1},{0,1,0},{0,-1,0},{1,0,0},{-1,0,0}}; char map[12][12][12]; struct node{ int x; int y; int z; }p,next; void bfs(int x,int y,int z) { queue<node>q; p.x=x; p.y=y; p.z=z; q.push(p); vis[z][x][y]=0; while(!q.empty()) { p=q.front(); q.pop(); if(p.x==ex&&p.y==ey&&p.z==ez) {return;} for(int i=0; i<6; i++) { int xx=p.x+dir[i][0]; int yy=p.y+dir[i][1]; int zz=p.z+dir[i][2]; if(xx>=0&&yy>=0&&zz>=0&&xx<n&&yy<n&&zz<n&&vis[zz][xx][yy]==inf&&map[zz][xx][yy]=='O') { vis[zz][xx][yy]=vis[p.z][p.x][p.y]+1; next.x=xx; next.y=yy; next.z=zz; q.push(next); } } } } int main() { char start[8],end[8]; int sx,sy,sz; while(scanf("%s %d",start,&n)!=EOF) { getchar(); memset(vis,inf,sizeof(vis)); for(int i=0; i<n; i++) { for(int j=0; j<n; j++) { scanf("%s",map[i][j]); } } scanf("%d%d%d%d%d%d",&sx,&sy,&sz,&ex,&ey,&ez); getchar(); scanf("%s",end); getchar(); bfs(sx,sy,sz); if(vis[ez][ex][ey]==inf) printf("NO ROUTE\n"); else printf("%d %d\n",n,vis[ez][ex][ey]); } return 0; }