題目描述：

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.

輸入：

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.

輸出：

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.

樣例輸入：

1 1
*
3 5
*@*@*
**@**
*@*@*
1 8
@@****@*
5 5
****@
*@@*@
*@**@
@@@*@
@@**@
0 0

樣例輸出：

0
1
2
2

Code:

#include <cstdio>using namespace std;char maze[101][101];   //保存地圖信息
bool mark[101][101];   //為圖上每個(gè)點(diǎn)設(shè)立一個(gè)狀態(tài)
int n,m;               //地圖大小n*m
int go[][2]={{1,0},{-1,0},{0,1},{0,-1},{1,1},{1,-1},{-1,-1},{-1,1}
};void DFS(int x,int y){for(int cnt=0;cnt<8;++cnt){int nx=x+go[cnt][0];int ny=y+go[cnt][1];if(nx<1||nx>n||ny<1||ny>m)    //該坐標(biāo)在地圖外continue;if(maze[nx][ny]=='*')continue;if(mark[nx][ny]==true)continue;mark[nx][ny]=true;DFS(nx,ny);}return;
}int main()
{while(scanf("%d%d",&n,&m)!=EOF){if(n==0&&m==0)break;for(int cnt=1;cnt<=n;++cnt){scanf("%s",maze[cnt]+1);}for(int cnt1=1;cnt1<=n;++cnt1){for(int cnt2=1;cnt2<=m;++cnt2){mark[cnt1][cnt2]=false;}}int ans=0;for(int cnt1=1;cnt1<=n;++cnt1){for(int cnt2=1;cnt2<=m;++cnt2){if(mark[cnt1][cnt2]==true)continue;if(maze[cnt1][cnt2]=='*')continue;DFS(cnt1,cnt2);++ans;}}printf("%d\n",ans);}return 0;
}/**************************************************************Problem: 1460User: lcyvinoLanguage: C++Result: AcceptedTime:10 msMemory:1040 kb
****************************************************************/

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轉(zhuǎn)載于:https://www.cnblogs.com/Murcielago/p/4167496.html

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