Red and Black
Tme Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18126 Accepted Submission(s): 11045
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
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#[email protected]#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
[email protected]
###.###
..#.#..
..#.#..
0 0
Sample Output
45
59
6
13
题意就是 找从@点开始出发,问所能到达的‘.‘点的个数(最开始理解错了意思,以为是一条路径的最大值,后面
才发现是可以到达的‘.‘点 ),直接DFS,
#include<iostream> #include<cstdio> #include<cmath> #include<map> #include<cstdlib> #include<vector> #include<set> #include<queue> #include<cstring> #include<string.h> #include<algorithm> #define INF 0x3f3f3f3f typedef long long ll; typedef unsigned long long LL; using namespace std; const int N = 1e6+10; const ll mod = 1e9+7; int t=0,flag=0; char s[26][26]; int visited[26][26]; int m,n; int _x[4]={0,1,-1,0}; int _y[4]={1,0,0,-1}; int ans; void DFS(int x,int y){ for(int t=0;t<4;t++){ int i=x+_x[t]; int j=y+_y[t]; if(i>=0&&j>=0&&visited[i][j]==0&&s[i][j]==‘.‘&&i<=n-1&&j<=m-1) { ans++; visited[i][j]=1; DFS(i,j); } } } int main(){ while(scanf("%d%d",&m,&n)!=EOF){ if(m==0&&n==0)break; memset(visited,0,sizeof(visited)); int ii,jj; int i,j; for(ii=0;ii<n;ii++){ for(jj=0;jj<m;jj++){ cin>>s[ii][jj]; if(s[ii][jj]==‘@‘){i=ii;j=jj;} } } ans=1; visited[i][j]=1; DFS(i,j); cout<<ans<<endl; } }