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.
InputThe 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)
OutputFor 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 Input6 9 ....#. .....# ...... ...... ...... ...... ...... #@...# .#..#. 11 9 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#[email protected]#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### [email protected] ###.### ..#.#.. ..#.#.. 0 0Sample Output
45 59 6 13
题目大意：“#”相当于不能走的陷阱或墙壁，“.”是可以走的路。从@点出发，统计所能到达的地点总数
1 //并查集解决
2 #include<iostream>
3 using namespace std;
4
5 const int h = 22;
6 char map[h][h];
7 int key[h*h];
8 int rrank[h*h];
9 int n,m,dx,dy;
10
11 int find(int a){
12 return a==key[a]? a : key[a]=find(key[a]);
13 }
14
15 void key_union(int a,int c){
16 int fa = find(a);
17 int fc = find(c);
18 if(rrank[fa]>rrank[fc])
19 key[fc] = fa;
20 else{
21 key[fa] = fc;
22 if(rrank[fa]==rrank[fc])
23 rrank[fc]++;
24 }
25 }
26
27 int num(int a){
28 int k = find(a);
29 int ans = 0;
30 for(int i=1;i<=m;i++)
31 for(int j=1;j<=n;j++)
32 if(find(i*n+j)==k)
33 ans++;
34
35 return ans;
36 }
37
38 int main()
39 {
40 while(scanf("%d %d",&n,&m)!=EOF){
41 if(n==0&&m==0) break;
42 for(int i=1;i<=m;i++){
43 cin.get();
44 for(int j=1;j<=n;j++){
45 scanf("%c",&map[i][j]);
46 if(map[i][j]!=‘#‘) key[i*n+j] = i*n+j;
47 else key[i*n+j] = 0;
48 if(map[i][j]==‘@‘){//找到@的坐标
49 dx = i;
50 dy = j;
51 map[i][j] = ‘.‘;
52 }
53 }
54 }
55
56 for(int i=1;i<m;i++){
57 for(int j=1;j<n;j++){
58 if(key[i*n+j]){
59 if(key[i*n+j+1])
60 key_union(i*n+j,i*n+j+1);
61 if(key[i*n+n+j])
62 key_union(i*n+n+j,i*n+j);
63 }
64 }
65 if(key[i*n+n])
66 if(key[i*n+2*n])
67 key_union(i*n+2*n,i*n+n);
68 }
69 for(int i=1;i<n;i++)
70 if(key[m*n+i])
71 if(key[m*n+i+1])
72 key_union(m*n+i,m*n+i+1);
73
74 int ans = num(dx*n+dy);
75 printf("%d\n",ans);
76 }
77 }
1 //DFS解决
2 #include<iostream>
3 using namespace std;
4
5 int mov[4][2] = {-1,0,1,0,0,-1,0,1};
6 int sum,w,h;
7 char s[21][21];
8
9 void dfs(int x,int y){
10 sum++;//计数
11 s[x][y] = ‘#‘;
12 for(int i=0;i<4;++i){//四个方向前进
13 int tx = x+mov[i][0];
14 int ty = y+mov[i][1];
15
16 if(s[tx][ty]==‘.‘ && tx>=0 && tx<h && ty>=0 && ty<w)
17 dfs(tx,ty);//判断该点可行后进入dfs
18 }
19 }
20
21 int main()
22 {
23 int x,y;
24 while(scanf("%d %d",&w,&h)!=EOF){
25 if(w==0&&h==0) break;
26 for(int i=0;i<h;i++){
27 cin.get();
28 for(int j=0;j<w;j++){
29 scanf("%c",&s[i][j]);
30 if(s[i][j]==‘@‘){//起点
31 x = i;
32 y = j;
33 }
34 }
35 }
36 sum = 0;
37 dfs(x,y);
38 printf("%d\n",sum);
39 }
40 return 0;
41 }
1 //BFS解决
2 #include<bits/stdc++.h>
3 using namespace std;
4
5 char room[23][23];
6 int dir[4][2] = { //左上角的坐标是(0,0)
7 {-1, 0}, //向左
8 {0, -1}, //向上
9 {1, 0}, //向右
10 {0, -1} //向下
11 };
12
13 int Wx, Hy, num;
14 #define check(x, y)(x<Wx && x>=0 && y>=0 && y<Hy) //是否在room中
15 struct node{int x, y};
16
17 void BFS(int dx, int dy){
18 num = 1;
19 queue<node> q;
20 node start, next;
21 start.x = dx;
22 start.y = dy;
23 q.push(start);//插入队列
24
25 while(!q.empty()){//直到队列为空
26 start = q.front();//取队首元素，即此轮循环的出发点
27 q.pop();//删除队首元素（以取出）
28
29 for(int i=0; i<4; i++){//往左上右下四个方向逐一搜索
30 next.x = start.x + dir[i][0];
31 next.y = start.y + dir[i][1];
32 if(check(next.x, next.y) && room[next.x][next.y]==‘.‘){
33 room[next.x][next.y] = ‘#‘;//标记已经走过
34 num ++;//计数
35 q.push(next);//判断此点可行之后，插入队列，待循环判断
36 }
37 }
38 }
39 }
40
41 int main(){
42 int x, y, dx, dy;
43 while(~scanf("%d %d",&Wx, &Hy)){
44 if(Wx==0 && Hy==0)
45 break;
46 for(y=0; y<Hy; y++){
47 for(x=0; x<Wx; x++){
48 scanf("%d",&room[x][y]);
49 if(room[x][y] == ‘@‘){//找到起点坐标
50 dx = x;
51 dy = y;
52 }
53 }
54 }
55 num = 0;//初始化
56 BFS(dx, dy);
57 printf("%d\n",num);
58 }
59 return 0;
60 }
原文地址：https://www.cnblogs.com/0424lrn/p/12230508.html