【POJ2251】Dungeon Master
【摘要】
problem
solution
codes
#include<iostream>
#include<queue>
#include<cstring>
using...
problem
solution
codes
#include<iostream>
#include<queue>
#include<cstring>
using namespace std;
int n, m, o, a[40][40][40], vis[40][40][40];
int sx, sy, sz, ex, ey, ez, ans;
const int dx[] = {1,-1,0,0,0,0};
const int dy[] = {0,0,1,-1,0,0};
const int dz[] = {0,0,0,0,1,-1};
struct node{
int x, y, z, step;
node(int x = 0, int y = 0, int z = 0, int step = 0):x(x),y(y),z(z),step(step){}
};
void bfs(){
queue<node>q;
q.push(node(sx,sy,sz,0));
a[sx][sy][sz] = 1;
//vis[sx][sy][sz] = 1;
while(!q.empty()){
node t = q.front(); q.pop();
if(t.x==ex && t.y==ey && t.z==ez){
ans = t.step;
}
for(int i = 0; i < 6; i++){
int nx = t.x+dx[i], ny = t.y+dy[i], nz = t.z+dz[i];
if(nx>0 && nx<=n && ny>0 && ny<=m && nz>0 && nz<=o && !a[nx][ny][nz]){
q.push(node(nx,ny,nz,t.step+1));
a[nx][ny][nz] = 1;
//vis[nx][ny][nz] = 1;
}
}
}
}
int main(){
while(cin>>n>>m>>o &&n){
ans = 0;
memset(a,0,sizeof(a));
//memset(vis,0,sizeof(vis));
for(int i = 1; i <= n; i++){
for(int j = 1; j <= m; j++){
for(int k = 1; k <= o; k++){
char ch; cin>>ch;
if(ch == '#')a[i][j][k] = 1;
if(ch == 'S'){ sx = i; sy = j; sz = k;}
if(ch == 'E'){ ex = i; ey = j; ez = k;}
}
}
}
bfs();
if(ans)cout<<"Escaped in "<<ans<<" minute(s).\n";
else cout<<"Trapped!\n";
}
return 0;
}
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文章来源: gwj1314.blog.csdn.net,作者:小哈里,版权归原作者所有,如需转载,请联系作者。
原文链接:gwj1314.blog.csdn.net/article/details/80461098
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