标签：miss character open tom eof use log color black
Red and Black |
| Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) |
| Total Submission(s): 118 Accepted Submission(s): 92 |
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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. |
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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) |
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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).
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Sample Input
6 9 ....#. .....# ...... ...... ...... ...... ...... #@...# .#..#. 11 9 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#..@#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### ...@... ###.### ..#.#.. ..#.#.. 0 0 |
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Sample Output
45 59 6 13 |
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Source
Asia 2004, Ehime (Japan), Japan Domestic
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Recommend
Eddy
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#include<bits/stdc++.h> #define N 25 using namespace std; int vis[N][N]; char mapn[N][N]; int dir[4][2]={ {1,0}, {-1,0}, {0,1}, {0,-1}, }; int cur=0; int n,m; bool ok(int x,int y) { if(x<0||x>=m||y<0||y>=n||vis[x][y]||mapn[x][y]==‘#‘) return false; return true; } void dfs(int x,int y) { cur++; for(int i=0;i<4;i++) { int fx=x+dir[i][0]; int fy=y+dir[i][1]; if(ok(fx,fy)) { vis[fx][fy]=1; dfs(fx,fy); } } } int main() { //freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin); while(scanf("%d%d",&n,&m)!=EOF&&n&&m) { memset(vis,0,sizeof vis); int x,y; getchar(); for(int i=0;i<m;i++) { scanf("%s",mapn[i]); //cout<<mapn[i]<<endl; for(int j=0;j<n;j++) { if(mapn[i][j]==‘@‘) { x=i; y=j; } //cout<<mapn[i][j]; } //cout<<endl; } cur=0; vis[x][y]=1; dfs(x,y); printf("%d\n",cur); } return 0; }
标签：miss character open tom eof use log color black
原文地址：http://www.cnblogs.com/wuwangchuxin0924/p/6041178.html
