<span style="color:#330099;">/* B - 广搜/深搜 基础 Time Limit:1000MS Memory Limit:30000KB 64bit IO Format:%I64d & %I64u Submit Status 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) The end of the input is indicated by a line consisting of two zeros. 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 By Grant Yuan 2014.7.12 */ #include<iostream> #include<stdio.h> #include<string.h> #include<queue> #include<cstdio> using namespace std; bool flag[21][21]; char a[21][21]; int n,m; int s1,f1; typedef struct{ int x; int y; }node; int next[4][2]={1,0,0,1,-1,0,0,-1}; int sum; queue<node>q; bool can(int cc,int dd) { if(cc>=0&&cc<m&&dd>=0&&dd<n&&flag[cc][dd]==0) return 1; return 0; } void slove() { int c,d,cc,dd; node q1; while(!q.empty()){ c=q.front().x;d=q.front().y; for(int i=0;i<4;i++) { cc=c+next[i][0]; dd=d+next[i][1]; if(can(cc,dd)) { q1.x=cc; q1.y=dd; q.push(q1); flag[cc][dd]=1; sum++; } } q.pop(); } } int main() { node q1; while(1){ sum=1; memset(flag,0,sizeof(flag)); cin>>n>>m; if(n==0&&m==0) break; for(int i=0;i<m;i++){ scanf("%s",&a[i]);} for(int i=0;i<m;i++) for(int j=0;j<n;j++) {if(a[i][j]=='#') flag[i][j]=1; if(a[i][j]=='@') s1=i,f1=j; } q1.x=s1; q1.y=f1; q.push(q1); flag[s1][f1]=1; slove(); cout<<sum<<endl; } return 0; } </span>
B广搜深搜
时间: 2024-10-03 00:57:54