Asteroids!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4775 Accepted Submission(s): 3070
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.
思路:題目主要是長,其實並不難,題中第一行輸入數n,接下來輸入n*n*n的三維圖,'O'表示能走,'X'不能走,倒數2、3行表示起點和終點的x、y、z座標,能抵達的話輸出n和最小步數,不能抵達則輸出"NO ROUTE",用三維的BFS即可。
#include<iostream>
#include<queue>
#include<cstring>
struct node
{
int x,y,z,step;
};
node st,ne,q,ed;
int to[6][3]={1,0,0 , 0,1,0 , 0,0,1 ,-1,0,0 , 0,-1,0 ,0,0,-1};
char map[30][30][30];
bool vis[30][30][30];
bool flag;
int n;
void bfs(void);
using namespace std;
int main(void)
{
char chs[100];
while(cin>>chs)
{
cin>>n;
for(int k=0;k<n;k++)
{
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
{
cin>>map[k][i][j];
}
}
}
cin>>st.x;
cin>>st.y;
cin>>st.z;
cin>>ed.x;
cin>>ed.y;
cin>>ed.z;
st.step=0;
ed.step=0;
cin>>chs;
memset(vis,0,sizeof(vis));
flag=false;
bfs();
if(!flag)
{
cout<<"NO ROUTE"<<endl;
}
}
return 0;
}
void bfs(void)
{
queue<node> que;
que.push(st);
while(!que.empty())
{
q=que.front();
que.pop();
if(q.x==ed.x && q.y==ed.y && q.z==ed.z)
{
cout<<n<<" "<<q.step<<endl;
flag=true;
return ;
}
for(int i=0;i<6;i++)
{
ne.x=q.x+to[i][0];
ne.y=q.y+to[i][1];
ne.z=q.z+to[i][2];
if(ne.x>=0 &&ne.x<n &&ne.y>=0 &&ne.y <n &&ne.z>=0 &&ne.z<n &&!vis[ne.z][ne.y][ne.x] && map[ne.z][ne.y][ne.x]=='O')
{
vis[ne.z][ne.y][ne.x]=true;
ne.step=q.step+1;
que.push(ne);
}
}
}
}