http://acm.hdu.edu.cn/showproblem.php?pid=1312
Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6074 Accepted Submission(s): 3863
Problem Description
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
The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.
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)
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
这题可用深搜也可用广搜,我用的广搜(BFS),第一次用广搜,好蛋疼,写了几个小时。。。终于搞定了。。
我一开始一直不对,是因为入队时没有处理好,让一个点重复入队几次,导致搜索次数大大增加,总是不对。。哭。。。
AC代码:
#include <iostream> #include <cstdio> #include <queue> #include <cstring> using namespace std; struct Node { int x,y; }; char map[25][25]; int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}}; int Bfs(Node p) { Node next; Node now; int cnt,i; queue<Node> q; q.push(p); cnt = 0; map[p.x][p.y] ='#'; while(!q.empty()) { now= q.front(); q.pop(); cnt++; for(i = 0; i < 4; i++) { next.x = now.x+dir[i][0]; next.y = now.y+dir[i][1]; if(map[next.x][next.y] == '.') { map[next.x][next.y] = '#'; //入队时标记已入队,防止重复入队 q.push(next); } } } return cnt; } int main() { Node p; int n,m,time,i,j,x,y; while(scanf("%d%d",&n,&m),n&&m) { memset(map,'#',sizeof(map)); for(i = 1; i <= m; i++) //输入 { getchar(); for(j = 1; j <= n; j++) { scanf("%c",&map[i][j]); if(map[i][j] == '@') { p.x = i; p.y = j; map[i][j] = '.'; } } } time = Bfs(p); printf("%d\n",time); } return 0; }