Total Submission(s): 6103 Accepted Submission(s): 3887
Write a program to count the number of black tiles which he can reach by repeating the moves described above.
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
'@' - a man on a black tile(appears exactly once in a data set)
6 9 ....#. .....# ...... ...... ...... ...... ...... #@...# .#..#. 11 9 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#..@#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### ...@... ###.### ..#.#.. ..#.#.. 0 0
45 59 6 13
解题思路:简单搜索题,深度优先搜索,广度优先搜索都可以。只要做到两点,题目即可AC。1,把所有能到达的地方走一遍,2,同一点计数不能有重复,走过后标记即可。
#include<cstdio> #include<queue> #include<string.h> using namespace std; char map[20][20]; int flag[20][20]; int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}}; int n,m; int sx,sy,maxstep=1; struct Node { int x; int y; }; void bfs() { queue<Node> Q; int x,y; int k; Node first,next; first.x=sx; first.y=sy; flag[sx][sy]=1; Q.push(first); while(!Q.empty()) { first=Q.front(); Q.pop(); for(k=0;k<4;k++) { x=first.x+dir[k][0]; y=first.y+dir[k][1]; if(x<0||x>=n||y<0||y>=m) continue; if(map[x][y]=='.'&&flag[x][y]==0) { next.x=x; next.y=y; maxstep++; Q.push(next); flag[x][y]=1; //记得标记啊 } } } } int main() { int i,j; while(scanf("%d%d",&m,&n)&&n||m) { memset(flag,0,sizeof(flag)); maxstep=1; for(i=0;i<n;i++) { scanf("%s",map[i]); for(j=0;j<m;j++) if(map[i][j]=='@') { sx=i; sy=j; } } bfs(); printf("%d\n",maxstep); } return 0; }