Red and Black(搜索题目)
Red and Black
Time Limit: 1000MSMemory Limit: 32768KB64bit IO Format: %I64d & %I64u
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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
Source
Asia 2004, Ehime (Japan), Japan Domestic
简单的搜索问题,根据题意,可以重复行走,求能到达的最多的...的数量,刚开始拿过题来以为是宽搜,因为做了几个宽搜的题目,形成思维定视了,后来一看到重复行走就想到了深搜。
#include <cstdio> #include <iostream> #include <cmath> #include <queue> #include <algorithm> using namespace std; int dx[4]={-1,0,1,0}; int dy[4]={0,-1,0,1}; int map[40][40]; int m,n; struct node { int x; int y; int sept; }; int sum; int bfs(int a,int b) { if (map[a][b]=='.') { map[a][b]='&'; sum++; } for (int number1=0;number1<4;number1++) { int nx=a+dx[number1],ny=b+dy[number1]; if (map[nx][ny]=='.'&&nx>=0&&nx<n&&ny>=0&&ny<m) { bfs(nx,ny); } } } int main() { while(scanf("%d %d",&m,&n)!=EOF) { if (m==0&&n==0)break; int i,j; getchar(); for (int number1=0;number1<n;number1++) { for (int number2=0;number2<m;number2++) { scanf("%c",&map[number1][number2]); if (map[number1][number2]=='@') { i=number1; j=number2; } } getchar(); } sum=1; bfs(i,j); printf("%d\n",sum); } return 0; }