标签:des style http io ar color os sp for
Time Limit: 1000MS | Memory Limit: 30000K | |
Total Submissions: 23194 | Accepted: 12513 |
Description
Input
Output
Sample Input
6 9 ....#. .....# ...... ...... ...... ...... ...... #@...# .#..#. 11 9 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#..@#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### ...@... ###.### ..#.#.. ..#.#.. 0 0
Sample Output
45 59 6 13题目翻译:给出一个m*n的图,然后‘.’代表可以走的路线,‘@’代表起点,‘#’代表墙,不能走,
求:算上起点,一共有多少格子可以走?
解题思路:dfs搜索!
#include<cstdio> int m,n,x1,y1,cot; char G[22][22]; int mov[][2]={0, 1, 0, -1, -1, 0, 1, 0}; void dfs(int x,int y){ G[x][y]='#'; int i,X,Y; for(i=0;i<4;i++){ X=x+mov[i][0]; Y=y+mov[i][1]; if(G[X][Y]!='#'&&X>=0&&X<n&&Y>=0&&Y<m){ cot++; dfs(X,Y); } } } int main(){ int i,j; while(scanf("%d%d",&m,&n),m|n){ for(i=0;i<n;++i) scanf("%s",G[i]); for(i=0;i<n;++i) for(j=0;j<m;++j) if(G[i][j]=='@'){ x1=i;y1=j; } cot=1; dfs(x1,y1); printf("%d\n",cot); } return 0; }
标签:des style http io ar color os sp for
原文地址:http://blog.csdn.net/hpuhjl/article/details/41514027