HDU 1312 Red and Black(最简单也是最经典的搜索)
传送门:
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): 25397 Accepted Submission(s): 15306
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.
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)
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
Recommend
分析:
只能上下左右四个方向走,问你可以走的块最多是多少?#不能走
小技巧:走过的地方字符就变为#
先用dfs写一下,有时间再用bfs写
code:
#include<bits/stdc++.h> using namespace std; #define max_v 25 char G[max_v][max_v]; int n,m; int sx,sy; int step; int dir[4][2]={0,1,1,0,0,-1,-1,0}; void dfs(int x,int y) { int xx,yy; for(int i=0;i<4;i++) { xx=x+dir[i][0]; yy=y+dir[i][1]; if(xx>=0&&xx<n&&yy>=0&&yy<m&&G[xx][yy]!='#') { step++; G[xx][yy]='#'; dfs(xx,yy); } } } int main() { while(~scanf("%d %d",&m,&n)) { if(n==0&&m==0) break; getchar(); for(int i=0;i<n;i++) { for(int j=0;j<m;j++) { cin>>G[i][j]; if(G[i][j]=='@') { sx=i; sy=j; } } } step=1; G[sx][sy]='#'; dfs(sx,sy); cout<<step<<endl; } return 0; }
心之所向,素履以往
分类:
ACM
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