HDU 1241 Oil Deposits 題解
阿新 • • 發佈：2018-11-20
Oil Deposits
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 45017 Accepted Submission(s): 25972
Input The input file contains one or more grids. Each grid begins with a line containing m and n, the number of rows and columns in the grid, separated by a single space. If m = 0 it signals the end of the input; otherwise 1 <= m <= 100 and 1 <= n <= 100. Following this are m lines of n characters each (not counting the end-of-line characters). Each character corresponds to one plot, and is either `*', representing the absence of oil, or `@', representing an oil pocket.
Output For each grid, output the number of distinct oil deposits. Two different pockets are part of the same oil deposit if they are adjacent horizontally, vertically, or diagonally. An oil deposit will not contain more than 100 pockets.
Sample Input 1 1 * 3 5 *@*@* **@** *@*@* 1 8 @@****@* 5 5 ****@ *@@*@ *@**@ @@@*@ @@**@ 0 0
Sample Output 0 1 2 2
Source Mid-Central USA 1997
Recommend Eddy ------------------------------------------------------------------------------------------------------------------------------------------------------------------- 本題應為DFS的題目 但由於博主學DFS學的比較菜 所以博主沒有采用DFS 而是採用了多次BFS的方式 逐行逐列搜尋map 一旦找到一個@就向八個方向BFS一次 每次BFS的時候把@變為* 這樣搜完整片map 就結束 下面上我全是註釋 誰都看的懂的程式碼 -------------------------------------------------------------------------------------------------------------------------------------------------------------------
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 int dir[8][2]={{-1,0},{1,0},{0,-1},{0,1},{-1,-1},{-1,1},{1,1},{1,-1}};//此處應為八個方向 33 int n,m; 34 char a[101][101]; 35 struct node 36 { 37 int x,y; 38 }; 39 void bfs(int x,int y) 40 { 41 int i; 42 queue<node> q; 43 node g; 44 g.x=x;g.y=y; 45 q.push(g); 46 a[x][y]='*'; 47 while(!q.empty()) 48 { 49 node t=q.front(); 50 q.pop(); 51 for(i=0;i<8;i++)//八個方向尋找是否成片 52 { 53 int dx=t.x+dir[i][0]; 54 int dy=t.y+dir[i][1]; 55 if(dx>=0&&dy>=0&&dx<m&&dy<n&&a[dx][dy]=='@')//如果出現@，變為*(防止下面重複運算) 56 { 57 a[dx][dy]='*'; 58 g.x=dx;g.y=dy; 59 q.push(g); 60 } 61 } 62 } 63 } 64 int main() 65 { 66 int x; 67 while(scanf("%d%d",&m,&n)!=EOF) 68 { 69 if(m==0&&n==0) 70 return 0; 71 x=0;//初始化 72 for(int i=0;i<m;i++) 73 scanf("%s",a[i]); 74 for(int i=0;i<m;i++) 75 { 76 for(int j=0;j<n;j++) 77 { 78 if(a[i][j]=='@')//找到一個@就bfs一下是否成一片，並且消除它們 79 { 80 bfs(i,j); 81 x++; 82 } 83 } 84 } 85 printf("%d\n",x); 86 } 87 return 0; 88 }------------------------------------------------------------------------------------------------------------------------------------------------------------------- Notes:DFS的方法暫時還沒有想好，但博主認為這樣的方法更好 2018-11-20 02:11:33 Author:LanceYu