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