HDU 1240 Asteroids! 題解
阿新 • • 發佈：2018-11-16
Asteroids!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6174 Accepted Submission(s): 3876
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.
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
3 4
NO ROUTE
1 //Author:LanceYu 2 #include<iostream> 3 #include<string> 4 #include<cstring> 5 #include<cstdio> 6 #include<fstream> 7 #include<iosfwd> 8 #include<sstream> 9 #include<fstream> 10 #include<cwchar> 11 #include<iomanip> 12 #include<ostream> 13 #include<vector> 14 #include<cstdlib> 15 #include<queue> 16 #include<set> 17 #include<ctime> 18 #include<algorithm> 19 #include<complex> 20 #include<cmath> 21 #include<valarray> 22 #include<bitset> 23 #include<iterator> 24 #define ll long long 25 using namespace std; 26 const double clf=1e-8; 27 //const double e=2.718281828; 28 const double PI=3.141592653589793; 29 const int MMAX=2147483647; 30 //priority_queue<int>p; 31 //priority_queue<int,vector<int>,greater<int> >pq; 32 struct node 33 { 34 int x,y,z,step; 35 }; 36 char str[11]; 37 int n,vis[101][101][101]; 38 char map[101][101][101]; 39 int dir[6][3]={{-1,0,0},{1,0,0},{0,-1,0},{0,1,0},{0,0,-1},{0,0,1}};//三維，六個自由度，三個方向 40 int bfs(int a,int b,int c,int x,int y,int z) 41 { 42 int i; 43 queue<node> q; 44 q.push(node{a,b,c,0}); 45 vis[a][b][c]=1; 46 while(!q.empty()) 47 { 48 node t=q.front(); 49 q.pop(); 50 if(t.x==x&&t.y==y&&t.z==z) 51 return t.step; 52 for(i=0;i<6;i++) 53 { 54 int dx=t.x+dir[i][0]; 55 int dy=t.y+dir[i][1]; 56 int dz=t.z+dir[i][2]; 57 if(dx>=0&&dy>=0&&dz>=0&&dx<n&&dy<n&&dz<n&&!vis[dx][dy][dz]&&map[dx][dy][dz]=='O')//三維bfs 58 { 59 vis[dx][dy][dz]=1; 60 q.push(node{dx,dy,dz,t.step+1}); 61 } 62 } 63 } 64 return -1; 65 } 66 int main() 67 { 68 int a,b,c,x,y,z; 69 while(scanf("%s%d",str,&n)!=EOF)//吸收start 70 { 71 getchar(); 72 memset(vis,0,sizeof(vis)); 73 for(int i=0;i<n;i++) 74 { 75 for(int j=0;j<n;j++) 76 { 77 scanf("%s",map[i][j]);//輸入map 78 } 79 } 80 scanf("%d%d%d%d%d%d",&a,&b,&c,&x,&y,&z);//輸入起點和終點 81 scanf("%s",str);//吸收掉end 82 int ans=bfs(a,b,c,x,y,z); 83 if(ans==-1) 84 printf("NO ROUTE\n"); 85 else 86 printf("%d %d\n",n,ans); 87 } 88 return 0; 89 }
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
然後恭喜你
第二組樣例
都過不了⊙﹏⊙∥
原因很簡單
題意說的n是按照層數排列的
而這裡的層數是z
而輸入的時候x作為深度
顯然就與題意不符了
這裡有兩種改法
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
第一種:
1 //Author:LanceYu 2 #include<iostream> 3 #include<string> 4 #include<cstring> 5 #include<cstdio> 6 #include<fstream> 7 #include<iosfwd> 8 #include<sstream> 9 #include<fstream> 10 #include<cwchar> 11 #include<iomanip> 12 #include<ostream> 13 #include<vector> 14 #include<cstdlib> 15 #include<queue> 16 #include<set> 17 #include<ctime> 18 #include<algorithm> 19 #include<complex> 20 #include<cmath> 21 #include<valarray> 22 #include<bitset> 23 #include<iterator> 24 #define ll long long 25 using namespace std; 26 const double clf=1e-8; 27 //const double e=2.718281828; 28 const double PI=3.141592653589793; 29 const int MMAX=2147483647; 30 //priority_queue<int>p; 31 //priority_queue<int,vector<int>,greater<int> >pq; 32 struct node 33 { 34 int x,y,z,step; 35 }; 36 char str[11]; 37 int n,vis[101][101][101]; 38 char map[101][101][101]; 39 int dir[6][3]={{-1,0,0},{1,0,0},{0,-1,0},{0,1,0},{0,0,-1},{0,0,1}};//六個自由度，三個方向 40 int bfs(int a,int b,int c,int x,int y,int z) 41 { 42 int i; 43 queue<node> q; 44 q.push(node{a,b,c,0}); 45 vis[a][b][c]=1; 46 while(!q.empty()) 47 { 48 node t=q.front(); 49 q.pop(); 50 if(t.x==x&&t.y==y&&t.z==z) 51 return t.step; 52 for(i=0;i<6;i++) 53 { 54 int dx=t.x+dir[i][0]; 55 int dy=t.y+dir[i][1]; 56 int dz=t.z+dir[i][2]; 57 if(dx>=0&&dy>=0&&dz>=0&&dx<n&&dy<n&&dz<n&&!vis[dx][dy][dz]&&map[dx][dy][dz]=='O')//三維bfs 58 { 59 vis[dx][dy][dz]=1; 60 q.push(node{dx,dy,dz,t.step+1}); 61 } 62 } 63 } 64 return -1; 65 } 66 int main() 67 { 68 int a,b,c,x,y,z; 69 while(scanf("%s%d",str,&n)!=EOF)//吸收掉start 70 { 71 getchar(); 72 memset(vis,0,sizeof(vis)); 73 for(int i=0;i<n;i++) 74 { 75 for(int j=0;j<n;j++) 76 { 77 scanf("%s",map[i][j]);//輸入map 78 } 79 } 80 scanf("%d%d%d%d%d%d",&c,&b,&a,&z,&y,&x);//注意輸入的方向是反過來的,層數是z,輸入的時候做了一個糾正 81 scanf("%s",str);//同理，吸收掉end 82 int ans=bfs(a,b,c,x,y,z); 83 if(ans==-1) 84 printf("NO ROUTE\n"); 85 else 86 printf("%d %d\n",n,ans); 87 } 88 return 0; 89 }
這裡在輸入起點與終點方面交換了一下位置
效果拔群
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
第二種:
1 //Author:LanceYu 2 #include<iostream> 3 #include<string> 4 #include<cstring> 5 #include<cstdio> 6 #include<fstream> 7 #include<iosfwd> 8 #include<sstream> 9 #include<fstream> 10 #include<cwchar> 11 #include<iomanip> 12 #include<ostream> 13 #include<vector> 14 #include<cstdlib> 15 #include<queue> 16 #include<set> 17 #include<ctime> 18 #include<algorithm> 19 #include<complex> 20 #include<cmath> 21 #include<valarray> 22 #include<bitset> 23 #include<iterator> 24 #define ll long long 25 using namespace std; 26 const double clf=1e-8; 27 //const double e=2.718281828; 28 const double PI=3.141592653589793; 29 const int MMAX=2147483647; 30 //priority_queue<int>p; 31 //priority_queue<int,vector<int>,greater<int> >pq; 32 struct node 33 { 34 int x,y,z,step; 35 }; 36 char str[11]; 37 int n,vis[101][101][101]; 38 char map[101][101][101]; 39 int dir[6][3]={{-1,0,0},{1,0,0},{0,-1,0},{0,1,0},{0,0,-1},{0,0,1}};//六個自由度，三個方向 40 int bfs(int a,int b,int c,int x,int y,int z) 41 { 42 int i; 43 queue<node> q; 44 q.push(node{a,b,c,0}); 45 vis[a][b][c]=1; 46 while(!q.empty()) 47 { 48 node t=q.front(); 49 q.pop(); 50 if(t.x==x&&t.y==y&&t.z==z) 51 return t.step; 52 for(i=0;i<6;i++) 53 { 54 int dx=t.x+dir[i][0]; 55 int dy=t.y+dir[i][1]; 56 int dz=t.z+dir[i][2]; 57 if(dx>=0&&dy>=0&&dz>=0&&dx<n&&dy<n&&dz<n&&!vis[dx][dy][dz]&&map[dx][dy][dz]=='O')//三維bfs 58 { 59 vis[dx][dy][dz]=1; 60 q.push(node{dx,dy,dz,t.step+1}); 61 } 62 } 63 } 64 return -1; 65 } 66 int main() 67 { 68 int a,b,c,x,y,z; 69 while(scanf("%s%d",str,&n)!=EOF)//吸收掉start 70 { 71 getchar(); 72 memset(vis,0,sizeof(vis)); 73 for(int i=0;i<n;i++) 74 { 75 for(int j=0;j<n;j++) 76 { 77 for(int k=0;k<n;k++) 78 cin>>map[k][j][i];//輸入map的時候直接按照需求輸入 79 } 80 } 81 scanf("%d%d%d%d%d%d",&a,&b,&c,&x,&y,&z);//輸入起點和終點 82 scanf("%s",str);//同理，吸收掉end 83 int ans=bfs(a,b,c,x,y,z); 84 if(ans==-1) 85 printf("NO ROUTE\n"); 86 else 87 printf("%d %d\n",n,ans); 88 } 89 return 0; 90 }
輸入時按照需求輸入
效果同樣拔群
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Notes:看清題目,看清題目,看清題目(重要的事情說三遍)
2018-11-16 03:24:04 Author:LanceYu