HDU 1979 Red and Black
題目:
There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can‘t move on red tiles, he can move only on black tiles.Write a program to count the number of black tiles which he can reach by repeating the moves described above.
Input
There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.
‘.‘ - a black tile
‘#‘ - a red tile
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The end of the input is indicated by a line consisting of two zeros.
Output
For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).Sample Input
6 9 ....#. .....# ...... ...... ...... ...... ...... #@...# .#..#. 11 9 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#..@#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### ...@... ###.### ..#.#.. ..#.#.. 0 0
Sample Output
45 59 6 13
題意描述:
輸入矩陣的大小W和H(均小於20)
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解題思路:
[email protected],隨後對其進行DFS搜索,下面的代碼實現的搜索有點模擬廣搜的意思。
代碼實現:
1 #include<stdio.h> 2 char map[30][30]; 3 int dfs(int x,int y); 4 int w,h; 5 int main() 6 { 7 int i,j,sx,sy; 8 while(scanf("%d%d",&w,&h),w+h != 0) 9 { 10 for(i=1;i<=h;i++) 11 { 12 for(j=1;j<=w;j++){ 13 scanf(" %c",&map[i][j]); 14 if(map[i][j]==‘@‘) 15 { sx=i;sy=j; } 16 } 17 getchar(); 18 } 19 printf("%d\n",dfs(sx,sy)); 20 } 21 return 0; 22 } 23 int dfs(int x,int y) 24 { 25 if(x<1 || x>h || y<1 || y>w) 26 return 0; 27 //如果進入不了dfs函數就是邊界問題,註意行數和列數就是x和y的範圍 28 if(map[x][y]==‘#‘) 29 return 0; 30 else 31 { 32 map[x][y]=‘#‘; 33 return 1+dfs(x-1,y)+dfs(x+1,y)+dfs(x,y-1)+dfs(x,y+1); 34 } 35 }
易錯分析:
1、如果搜索進入不了註意邊界的設置問題
HDU 1979 Red and Black