DFS 图的联通块 UVA-572 Oil Deposits 题目链接：[ UVA - 572 ]   The GeoSurvComp geologic survey company is responsible for detecting underground oil deposits. GeoSurvComp works with one large rectangular region of land at a time, and creates a grid that divides the land into numerous square plots. It then analyzes each plot separately, using sensing equipment to determine whether or not the plot contains oil.   A plot containing oil is called a pocket. If two pockets are adjacent, then they are part of the same oil deposit. Oil deposits can be quite large and may contain numerous pockets. Your job is to determine how many different oil deposits are contained in a grid. Input The input file contains one or more grids. Each grid begins with a line containing m and n, the number of rows and columns in the grid, separated by a single space. If m = 0 it signals the end of the input; otherwise 1 ≤ m ≤ 100 and 1 ≤ n ≤ 100. Following this are m lines of n characters each (not counting the end-of-line characters). Each character corresponds to one plot, and is either ‘*’, representing the absence of oil, or ‘@’, representing an oil pocket. Output For each grid, output the number of distinct oil deposits. Two different pockets are part of the same oil deposit if they are adjacent horizontally, vertically, or diagonally. An oil deposit will not contain more than 100 pockets. Sample Input 1 1 * 3 5 *@*@* **@** *@*@* 1 8 @@****@* 5 5 ****@ *@@*@ *@**@ @@@*@ @@**@ 0 0 Sample Output 0 1 2 2 题解 从每个"@"格子出发，递归遍历它周围的"@"格子。 每次访问就把他标记成"*"，即不通路（或者可以用数组表示是否访问过） #include<iostream> #include<string> #include<cstdio> #include<cstring> #include<queue> #include<map> #include<set> #include<algorithm> #include<cmath> #include<cstdlib> using namespace std; int m, n; char garden[105][105]; void dfs(int x, int y) { //将当前点取消标记，避免重复查找 garden[x][y] = '*'; //遍历周围的八个点 for (int dx = -1; dx <= 1; dx++){ for (int dy = -1; dy <= 1;dy++){ int nx = x + dx; int ny = y + dy; if(0<=nx && nx<n && 0<=ny && ny<m && garden[nx][ny]=='@') dfs(nx, ny); } } } int main(void) { while(~scanf("%d%d", &n, &m) && m+n){ getchar(); //吸收两数字后的换行符 memset(garden, 0, sizeof(garden)); for (int i = 0; i < n;i++){ for (int j = 0; j < m;j++) scanf("%c", &garden[i][j]); getchar(); //吸收每次输入一行后的换行符 } int sum = 0; for (int i = 0; i < n; i++){ for (int j = 0; j < m; j++){ if (garden[i][j] == '@'){ dfs(i, j); sum++; } } } cout << sum << endl; } system("pause"); return 0; }

DFS POJ-2386 Lake Counting [POJ - 2386] Due to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle of N x M (1 <= N <= 100; 1 <= M <= 100) squares. Each square contains either water ('W') or dry land ('.'). Farmer John would like to figure out how many ponds have formed in his field. A pond is a connected set of squares with water in them, where a square is considered adjacent to all eight of its neighbors. Given a diagram of Farmer John's field, determine how many ponds he has. Input Line 1: Two space-separated integers: N and M Lines 2..N+1: M characters per line representing one row of Farmer John's field. Each character is either 'W' or '.'. The characters do not have spaces between them. Output Line 1: The number of ponds in Farmer John's field. Sample Input 10 12 W........WW. .WWW.....WWW ....WW...WW. .........WW. .........W.. ..W......W.. .W.W.....WW. W.W.W.....W. .W.W......W. ..W.......W. Sample Output 3 Hint OUTPUT DETAILS: There are three ponds: one in the upper left, one in the lower left,and one along the right side. 题解 #include<iostream> #include<string> #include<cstdio> #include<cstring> #include<queue> #include<map> #include<set> #include<algorithm> #include<cmath> #include<cstdlib> using namespace std; int m, n; char garden[105][105]; void dfs(int x, int y) { //将当前点取消标记，避免重复查找 garden[x][y] = '.'; //遍历周围的八个点 for (int dx = -1; dx <= 1; dx++){ for (int dy = -1; dy <= 1;dy++){ int nx = x + dx; int ny = y + dy; if(0<=nx && nx<n && 0<=ny && ny<m && garden[nx][ny]=='W') dfs(nx, ny); } } } int main(void) { while(~scanf("%d%d", &n,&m)){ getchar(); //吸收两数字后的换行符 memset(garden, 0, sizeof(garden)); for (int i = 0; i < n;i++){ for (int j = 0; j < m;j++) scanf("%c", &garden[i][j]); getchar(); //吸收每次输入一行后的换行符 } int sum = 0; for (int i = 0; i < n; i++){ for (int j = 0; j < m; j++){ if (garden[i][j] == 'W'){ dfs(i, j); sum++; } } } cout << sum << endl; } return 0; }

DFS 图的联通块 POJ-1979 Red and Black [ POJ - 1979]   Here 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 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#..@#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### ...@... ###.### ..#.#.. ..#.#.. 0 0 Sample Output 45 59 6 13 题解 #include<iostream> #include<string> #include<cstdio> #include<cstring> #include<queue> #include<map> #include<set> #include<algorithm> #include<cmath> #include<cstdlib> using namespace std; int m, n, sum; char garden[105][105]; int d[4][2] = { {0, -1}, {1, 0}, {0, 1}, {-1, 0}, }; void dfs(int x, int y) { garden[x][y] = '#'; sum++; for (int i = 0; i < 4; i++){ int nx = x + d[i][1]; int ny = y + d[i][0]; if(0<=nx && nx<n && 0<=ny && ny<m && garden[nx][ny]=='.') dfs(nx, ny); } } int main(void) { while(~scanf("%d%d", &m, &n) && m+n){ getchar(); memset(garden, 0, sizeof(garden)); for (int i = 0; i < n; i++){ for (int j = 0; j < m; j++) scanf("%c", &garden[i][j]); getchar(); } sum = 0; for (int i = 0; i < n; i++) for (int j = 0; j < m; j++) if (garden[i][j] == '@') dfs(i, j); cout << sum << endl; } return 0; }

DFS 图的联通块 AOJ-0118 Property Distribution [ Aizu - 0118 ] 这道题类似于联通图问题，将联通的归成一个，数一下总共有几个即可。 因为题目告诉不会有空格，所以排除标记用空格表示即可。 Sample Input 10 10 ####*****@ @#@@@@#*#* @##***@@@* #****#*@** ##@*#@@*## *@@@@*@@@# ***#@*@##* *@@@*@@##@ *@*#*@##** @****#@@#@ 0 0 Sample Output 33 题解 #include<iostream> #include<string> #include<cstdio> #include<cstring> #include<queue> #include<map> #include<set> #include<algorithm> #include<cmath> #include<cstdlib> using namespace std; int m, n; char garden[105][105]; int d[4][2] = { {0, -1}, {1, 0}, {0, 1}, {-1, 0}, }; void dfs(int x, int y, char c) { if(garden[x][y] == c){ garden[x][y] = ' '; for (int i = 0; i < 4; i++){ int nx = x + d[i][1]; int ny = y + d[i][0]; if (0 <= nx && nx < n && 0 <= ny && ny < m && garden[nx][ny] != ' ') dfs(nx, ny, c); } } } int main(void) { while(~scanf("%d%d", &n, &m) && m+n){ getchar(); memset(garden, 0, sizeof(garden)); for (int i = 0; i < n; i++){ for (int j = 0; j < m; j++) scanf("%c", &garden[i][j]); getchar(); } int sum = 0; for (int i = 0; i < n; i++){ for (int j = 0; j < m; j++){ if (garden[i][j] != ' '){ dfs(i, j, garden[i][j]); sum++; } } } cout << sum << endl; } return 0; }