UVA - 572 Oil Deposits（DFS和BFS两种解法）

  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 containingm andn, the number of rows and columns in the grid, separated by a single space. Ifm = 0 it signals the end of the input; otherwise and. Following this arem 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 

0122

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0continue;}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;}