HDU 1312 Red and Black(DFS,板子题,详解,零基础教你代码实现DFS)
2017-04-07 21:57
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Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 19731 Accepted Submission(s): 11991[align=left]Problem Description[/align]
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.
[align=left]Input[/align]
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)
[align=left]Output[/align]
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).
[align=left]Sample Input[/align]
6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0
[align=left]Sample Output[/align]
45
59
6
13
[align=left]Source[/align]
Asia 2004, Ehime (Japan), Japan Domestic
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1312
分析:看到ZERO已经写了DFS,弱鸡也得补上此题,来一发!感谢梅大大的支持,代码的模版来自梅大大,我只是对主要步骤做了注释而已!
下面给出AC代码:
#include <bits/stdc++.h> using namespace std; int _x[4]={1,0,-1,0}; int _y[4]={0,-1,0,1};//相当于从第一象限开始顺时针转一周,对应的坐标轴上的点分别为(1,0),(0,-1),(-1,0),(0,1),就是数学意义上的方向向量 char str[30][30]; bool flag[30][30];//去判断点是否被标记过,DFS搜索一遍,被标记过记改点为1! int n,m,ans; void DFS(int x,int y) { for(int ii=0;ii<4;ii++) { int i=x+_x[ii]; int j=y+_y[ii]; if(i<n&&j<m&&i>=0&&j>=0&&flag[i][j]==0&&str[i][j]=='.')//边界条件 { ans++; flag[i][j]=1; DFS(i,j); } } } int main() { while(scanf("%d%d",&m,&n)!=EOF) { if(m==0&&n==0) break; int x,y; memset(flag,0,sizeof(flag)); for(int i=0;i<n;i++) scanf("%s",str[i]); for(int i=0;i<n;i++) { for(int j=0;j<m;j++) { if(str[i][j]=='@')//找到起始点 { x=i; y=j; } } } ans=0; flag[x][y]=1; DFS(x,y); printf("%d\n",ans+1);//第一个位置也算一个,所以+1 } return 0; }
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