UVA - 572 Oil Deposits（DFS和BFS两种解法）

最新推荐文章于 2020-03-03 09:50:26 发布

原创最新推荐文章于 2020-03-03 09:50:26 发布 · 931 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#Uva #Oil Deposits

dfs or bfs 专栏收录该内容

50 篇文章

订阅专栏

地质勘探公司使用感应设备在矩形区域的土地上进行油气勘探，将土地划分为多个正方形地块，并分别分析每个地块是否存在石油。一块包含石油的地块称为口袋，相邻的口袋属于同一油藏。任务是确定网格中不同油藏的数量。

Oil Deposits

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 $1 \le m \le 100$ and $1 \le n \le 100$ . 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

题意：

给你一个二维数组，要你算出油田有几块。

其中'@'是有油，'*'是没有油

解析：

这是一道基础的搜索题目

以下有两种解法：

dfs解法：如果遇到'@'就cut++，并把'@'周围的'@'都变为'*'

#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
const int N = 110;
char map[N][N];
int m,n;
void dfs(int x,int y) {	
	if(x >= 0 && x <m && y >= 0 && y < n) {
		if(map[x][y] == '@') {
			map[x][y] = '*';
			for(int dr = -1; dr <= 1; dr++) {
				for(int dc = -1; dc <= 1; dc++) {
					if(!dr && !dc)
						continue;
					dfs(x+dr,y+dc);
				}
			}
		}
	}
}
int main() {
	int cut;
	while(cin >> m >> n && (m || n)) {
		getchar();
		for(int i = 0; i < m; i++) {
			for(int j = 0; j < n; j++) {
				cin >> map[i][j];
			}
		}
		cut = 0;
		for(int i = 0; i < m; i++) {
			for(int j = 0; j < n; j++) {
				if(map[i][j] == '@') {
					cut++;
					dfs(i,j);
				}
			}
		}
		cout<<cut<<endl;
	}
	return 0;
}

bfs解法：使用floodfill算法

#include <stdio.h>
#include <string.h>
#include <queue>
using namespace std;

const int N = 110;
char grid[N][N];
int vis[N][N];
int h,w;

void bfs(int r ,int c) {	
	queue<int> q;	
	q.push(r);
	q.push(c);	
	while( !q.empty() ) {
		r = q.front();
		q.pop();
		c = q.front();
		q.pop();	
		for(int dr = -1; dr <= 1; dr++) {
			for(int dc = -1; dc <= 1; dc++) {
				if(!dr && !dc) { //dr和dc不能同时为0
					continue;
				}
				if(r + dr < 0 || r + dr >= h || c + dc < 0 || c + dc >= w) { //越界
					continue;
				}
				if( grid[r+dr][c+dc] == '@' && !vis[r+dr][c+dc]) {
					vis[r+dr][c+dc] = 1;
					q.push(r+dr);
					q.push(c+dc);
				}
			}
		}
	}
}

int main() {
	int cnt;
	while(scanf("%d%d",&h,&w) != EOF && ( h || w)) {
		memset(grid,0,sizeof(grid));
		memset(vis,0,sizeof(vis));
		for(int i = 0; i < h; i++) {
			scanf("%s",grid[i]);
		}
		int r,c;
		cnt = 0;
		for(int i = 0; i < h; i++) {
			for(int j = 0; j < w; j++) {
				if(grid[i][j] == '@' && !vis[i][j]) {	
					cnt++;
					bfs(i,j);
				}
			}
		}
		printf("%d\n",cnt);
	}
	return 0;
}