Oil Deposits
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12461 Accepted Submission(s): 7245
Problem 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 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
本题是指@表示油点，*表示非油点，油点相邻则表示一个油点而不是多个，找出总油点个数。没啥技巧，就是八个方向搜索，同时利用并查集的效率更高
DFS搜索：
#include<bits/stdc++.h>
using namespace std;
const int N = 100 + 5;
char g[N][N];
bool book[N][N];
int n, m;
const int dx[] = { 0,0,1,1,1,-1,-1,-1 };
const int dy[] = { 1,-1,0,1,-1,0,1,-1 };
void dfs(int i, int j) {
book[i][j] = 1;
for (int k = 0; k < 8; k++){
int nx = i + dx[k], ny = j + dy[k];
if (nx >= 0 && nx < m && ny >= 0 && ny < n && book[nx][ny] == 0 && g[nx][ny] == '@')dfs(nx, ny);
}
}
int main() {
//freopen("in.txt", "r", stdin);
while (cin >> m >> n && n && m) {
memset(g, 0, sizeof(g));
memset(book, 0, sizeof(book));
for (int i = 0; i < m; i++)cin >> g[i]; int cnt = 0;
for (int i = 0; i < m; ++i)for (int j = 0; j < n; ++j)if (g[i][j] == '@' && !book[i][j]) { cnt++; dfs(i, j); }
cout << cnt << endl;
}
return 0;
}
并查集:
#include<bits/stdc++.h>
using namespace std;
char g[101][101];
int fa[101 * 101];
int x[] = { 0,0,1,-1,1,-1,1,-1 };
int y[] = { 1,-1,0,0,1,1,-1,-1 };
int n, m;
int id(int x, int y) {
return (x - 1)*m + y;
}
bool isok(int x, int y) {
if (0 < x&&x <= n && 0 < y&&y <= m)
return true;
return false;
}
int Find(int x) {
return fa[x] == x ? x : fa[x] = Find(fa[x]);
}
int vis[101 * 101];
int main() {
while (scanf("%d%d", &n, &m) != EOF && n + m) {
memset(vis, 0, sizeof vis);
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= m; ++j) {
fa[id(i, j)] = id(i, j);
}
}
for (int i = 1; i <= n; ++i)
scanf("%s", g[i] + 1);
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= m; ++j) {
if (g[i][j] == '*') continue;
for (int k = 0; k < 7; ++k) {
int xx = i + x[k], yy = j + y[k];
if (isok(xx, yy) == false || g[xx][yy] == '*') continue;
int f1 = Find(id(i, j)), f2 = Find(id(xx, yy));
if (f1 != f2) {
fa[f1] = f2;
}
}
}
}
int cnt = 0;
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= m; ++j) {
if (g[i][j] != '*'&&vis[Find(id(i, j))] == 0) {
cnt++;
vis[Find(id(i, j))] = 1;
}
}
}
cout << cnt << endl;
}
return 0;
}