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
思路 从起始点往4个方向开始遍历，但是最开始一直想dfs结束条件是什么？
void dfs（）{ if(??){ /* block code */ return ; } /* block code */ return; }这题dfs其实不需要if判断结束.
以下是我的代码
/************************************************************************* > File Name: hdu1312.cpp > Author:gens_ukiy > Mail: > Created Time: 2016年12月02日 星期五 13时35分18秒 ************************************************************************/ #include<iostream> #include<cstdio> #include<algorithm> #include<cmath> #include<cstdlib> #include<climits> #include<string> #include<cstring> #include<vector> #include<set> #include<list> #include<map> int dire[4][2]={ {0,1},{1,0},{0,-1},{-1,0} }; int dire2[8][2]={{-1,-1},{-1,0},{-1,1},{ 0,-1},{ 0,1},{ 1,-1},{ 1,0},{ 1,1}}; #define rep(i,a,b) for(int i=(a);i<=(b);(i++)) #define inf 0x3f3f3f #define ll long long #define pi acos(-1) using namespace std; char a[21][21]; int book[21][21]; int n,m; int cnt; void dfs(int x,int y){ if(book[x][y]==0){ cnt++; book[x][y]=1; rep(i,0,3){ int px=x+dire[i][0]; int py=y+dire[i][1]; if(px>=1&&px<=n&&py>=1&&py<=m&&book[px][py]==0&&a[px][py]!='#'){ dfs(px,py); } } } return ; } int main() { while(cin>>m>>n){ if(n==0&&m==0) break; memset(book,0,sizeof(book)); cnt=0; int sx,sy; rep(i,1,n) rep(j,1,m){ cin>>a[i][j]; if(a[i][j]=='@') sx=i,sy=j; } dfs(sx,sy); printf("%d\n",cnt); } return 0; }后来我又百度看了下大牛们的代码. 发现他们没有用book数组记录是否已经遍历,而是直接将当前字符修改为不可遍历字符,可以减少空间开辟.
/************************************************************************* > File Name: hdu1312_1.cpp > Author:gens_ukiy > Mail: > Created Time: 2016年12月02日 星期五 14时05分37秒 ************************************************************************/ #include<iostream> #include<cstdio> #include<algorithm> #include<cmath> #include<cstdlib> #include<climits> #include<string> #include<cstring> #include<vector> #include<set> #include<list> #include<map> int dire[4][2]={ {0,1},{1,0},{0,-1},{-1,0} }; int dire2[8][2]={{-1,-1},{-1,0},{-1,1},{ 0,-1},{ 0,1},{ 1,-1},{ 1,0},{ 1,1}}; #define rep(i,a,b) for(int i=(a);i<=(b);(i++)) #define inf 0x3f3f3f #define ll long long #define pi acos(-1) #define OnlineJudge using namespace std; char a[21][21]; int n,m; int cnt; void dfs(int x,int y){ cnt++;//能进入dfs，默认当前字符为'.' a[x][y]='#'; rep(i,0,3){ int px=x+dire[i][0]; int py=y+dire[i][1]; if(px>=1&&px<=n&&py>=1&&py<=m&&a[px][py]!='#'){ dfs(px,py); } } return; } int main() { #ifndef OnlineJudge freopen("in.txt","r",stdin); freopen("out.txt","w",stdout); #endif while(cin>>m>>n){ if(m==0&&n==0) break; int sx,sy; cnt=0; //每次都要初始化 rep(i,1,n) rep(j,1,m){ cin>>a[i][j]; if(a[i][j]=='@') sx=i,sy=j; //记录起始位置 } dfs(sx,sy); printf("%d\n",cnt); } return 0; }