Dungeon Master POJ - 2251(bfs)
程序员文章站
2022-06-12 10:14:20
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对于3维的,可以用结构体来储存,详细见下列代码。
样例可以过,不过能不能ac还不知道,疑似poj炸了,
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int INF = 0x3f3f3f3f;
int dx[] = {1, -1, 0, 0, 0, 0};
int dy[] = {0, 0, -1, 1, 0, 0};
int dz[] = {0, 0, 0, 0, 1, -1};
int bx, by, bz;
int ex, ey, ez;
const int maxn = 30 + 5;
class POS
{
public:
POS():x_(-1), y_(-1), z_(-1){ }
POS(int x, int y, int z):x_(x), y_(y), z_(z){ }
int getX(){ return x_;}
int getY(){ return y_;}
int getZ(){ return z_;}
private:
int x_, y_, z_;
};
typedef POS P;
struct Level
{
char maze[maxn][maxn];
int vis[maxn][maxn];
int d[maxn][maxn];
};
int L, R, C;
Level *h;
bool op(int x, int y, int z)
{
if(x >= 0 && x < R && y >= 0 && y < C && z >= 0 && z < L)
return true;
return false;
}
int bfs()
{
queue<P> pos;
pos.push(P(bx, by, bz));
while(!pos.empty())
{
P p = pos.front();
pos.pop();
if(p.getX() == ex && p.getY() == ey && p.getZ() == ez)
break;
h[p.getZ()].vis[p.getX()][p.getY()] = 1;
for(int i = 0; i < 6; i++)
{
int curx = p.getX() + dx[i], cury = p.getY() + dy[i], curz = p.getZ() + dz[i];
if(op(curx, cury, curz) && h[curz].maze[curx][cury] != '#' && !h[curz].vis[curx][cury])
{
h[curz].vis[curx][cury] = 1;
pos.push(P(curx, cury, curz));
h[curz].d[curx][cury] = h[p.getZ()].d[p.getX()][p.getY()] + 1;
}
}
}
return h[ez].d[ex][ey];
}
int main()
{
while(cin >> L >> R >> C && (L || R || C))
{
bx = by = bz = -1;
ex = ey = ez = -1;
h = new Level[L];
for(int i = 0; i < L; i++)
{
for(int j = 0; j < R; j++)
{
scanf("%s", h[i].maze[j]); //input
for(int k = 0; k < C ; k++)
{
h[i].vis[j][k] = 0;
if(h[i].maze[j][k] == 'S')
{
h[i].d[j][k] = 0;
bx = j;
by = k;
bz = i;
}
else
h[i].d[j][k] = INF;
if(h[i].maze[j][k] == 'E')
{
ex = j;
ey = k;
ez = i;
}
}
}
}
int ans = bfs();
if(ans == INF)
cout << "Trapped!" << endl;
else
cout << "Escaped in " << ans << " minute(s)." << endl;
delete []h;
}
}
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