Oil Deposits
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 14628 | Accepted: 7972 |
Description
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 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
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 方法1:并查集
收获：二维数组转一维数组公式：i*m(为列数）+j;
#include <cstdio> #include <iostream> #include <cstdlib> #include <algorithm> #include <ctime> #include <cmath> #include <string> #include <cstring> #include <stack> #include <queue> #include <list> #include <vector> #include <map> #include <set> using namespace std; const int INF=0x3f3f3f3f; const double eps=1e-10; const double PI=acos(-1.0); #define maxn 500 #define maxm 100000 int n, m; char mp[maxn][maxn]; int root[maxm]; bool solve(int x, int y) { if(x < 0 || x >= n || y < 0 || y >= m) return true; return false; } int find_root(int x) { if(x != root[x]) root[x] = find_root(root[x]); return root[x]; } void uni(int a, int b) { int x = find_root(a); int y = find_root(b); if(x != y) { root[y] = x; } } void judge(int x, int y) { for(int i = -1; i <= 1; i++) for(int j = -1; j <= 1; j++) { if(i != 0 || j != 0) { int dx = x + i; int dy = y + j; if(mp[dx][dy] == '@' && !solve(dx, dy)) { int p = x * m + y; int q = dx * m + dy; uni(p, q); } } } } int main() { while(~scanf("%d%d", &n, &m) && (n+m) != 0) { memset(root, -1, sizeof root); for(int i = 0; i < n; i++) scanf("%s", mp[i]); for(int i = 0; i < n; i++) { for(int j = 0; j < m; j++) { if(mp[i][j] == '@') root[i*m+j] = i * m +j; } } for(int i = 0; i < n; i++) for(int j = 0; j < m; j++) { if(mp[i][j] == '@') { judge(i, j); } } int ans = 0; for(int i = 0; i < n ; i++) for(int j = 0; j < m; j++) if(root[i*m+j] == i*m+j) ans++; printf("%d ", ans); } return 0; }
方法二：
DFS入门题
收获：做搜索的题目调试的时候可以用打印中间路径的方法来调试。
#include <cstdio> #include <iostream> #include <cstdlib> #include <algorithm> #include <ctime> #include <cmath> #include <string> #include <cstring> #include <stack> #include <queue> #include <list> #include <vector> #include <map> #include <set> using namespace std; const int INF=0x3f3f3f3f; const double eps=1e-10; const double PI=acos(-1.0); #define maxn 500 int n, m; char mp[maxn][maxn]; int vis[maxn][maxn]; bool solve(int x, int y) { if(x < 0 || x >= n || y < 0 || y >= m) return true; return false; } void dfs(int x, int y) { for(int i = -1; i <= 1; i++) for(int j = -1; j <= 1; j++) { if(i != 0 || j != 0) { int dx = x + i; int dy = y + j; if(!vis[dx][dy] && mp[dx][dy] == '@' && !solve(dx, dy)) { vis[dx][dy] = 1; dfs(dx, dy); } } } } int main() { while(~scanf("%d%d", &n, &m) && (n+m) != 0) { for(int i = 0; i < n; i++) scanf("%s", mp[i]); int cnt = 0; memset(vis, 0, sizeof vis); for(int i = 0; i < n; i++) for(int j = 0; j < m; j++) { if(mp[i][j] == '@' && !vis[i][j]) { vis[i][j] = 1; ++cnt; dfs(i, j); } } printf("%d ", cnt); } return 0; }