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HDU1241-Oil Deposits
Oil Deposits
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 9 Accepted Submission(s) : 6
Problem DescriptionThe 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.
InputThe 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.
OutputFor 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 Input1 1
*
3 5
*@*@*
**@**
*@*@*
1 8
@@****@*
5 5
****@
*@@*@
*@**@
@@@*@
@@**@
0 0
Sample Output0
1
2
2
題意：所給你一個地圖，有兩種元素分別為‘.’與@,然後要求你找出不相連的@的個數，@的鄰接與@的對角都算是相連的。
注意：第三組測試數據的"5 5“後有個空格，複製粘貼會出錯，建議手動輸入
//DFS
#include<stdio.h>
#include<string.h>
struct node
{
char V;
int flag;
}a[200][200];
int n,m;
void Fun(int x,int y)
{
if(x+1<n&&y-1>=0&&a[x+1][y-1].flag==0&&a[x+1][y-1].V=='@')
{
a[x+1][y-1].flag=1;
Fun(x+1,y-1);
}
if(x+1<n&&y+1<m&&a[x+1][y+1].flag==0&&a[x+1][y+1].V=='@')
{
a[x+1][y+1].flag=1;
Fun(x+1,y+1);
}
if(x-1>=0&&y-1>=0&&a[x-1][y-1].flag==0&&a[x-1][y-1].V=='@')
{
a[x-1][y-1].flag=1;
Fun(x-1,y-1);
}
if(x-1>=0&&y+1<m&&a[x-1][y+1].flag==0&&a[x-1][y+1].V=='@')
{
a[x-1][y+1].flag=1;
Fun(x-1,y+1);
}
if(x+1<n&&a[x+1][y].V=='@'&&a[x+1][y].flag==0)
{
a[x+1][y].flag=1;
Fun(x+1,y);
}
if(x-1>=0&&a[x-1][y].V=='@'&&a[x-1][y].flag==0)
{
a[x-1][y].flag=1;
Fun(x-1,y);
}
if(y+1<m&&a[x][y+1].V=='@'&&a[x][y+1].flag==0)
{
a[x][y+1].flag=1;
Fun(x,y+1);
}
if(y-1>=0&&a[x][y-1].V=='@'&&a[x][y-1].flag==0)
{
a[x][y-1].flag=1;
Fun(x,y-1);
}
return;
}
int main()
{
int i,j,sum;
while(scanf("%d %d",&n,&m),n!=0&&m!=0)
{
memset(a,0,sizeof(a));
for(i=0;i<n;i++)
{
getchar();
for(j=0;j<m;j++)
scanf("%c",&a[i][j].V);
}
sum=0;
for(i=0;i<n;i++)
{
for(j=0;j<m;j++)
{
if(a[i][j].V=='@'&&a[i][j].flag==0)
{
a[i][j].flag=1;
Fun(i,j);
sum++;
}
}
}
printf("%d\n",sum);
}
return 0;
}
最後更新：2017-04-03 12:56:23