hdoj 1241 【DFS】
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Oil Deposits
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
0122分析:简单的深搜,计算连着的油桶的数量。代码:Online Judge Online Exercise Online Teaching Online Contests Exercise Author F.A.QHand In HandOnline Acmers Forum | DiscussStatistical Charts Problem ArchiveRealtime Judge StatusAuthors Ranklist C/C++/Java Exams ACM StepsGo to JobContest LiveCastICPC@China Best Coder betaVIP | STD ContestsVirtual Contests DIY | Web-DIY betaRecent Contests 小破孩儿 Mail 0(0) Control Panel Sign Out 欢迎参加——每周六晚的BestCoder(有米!) View CodeProblem : 1241 ( Oil Deposits ) Judge Status : AcceptedRunId : 14334415 Language : G++ Author : xiaopohaierCode Render Status : Rendered By HDOJ G++ Code Render Version 0.01 Beta#include<stdio.h>#include<string.h>#include<algorithm>#include<stdlib.h>using namespace std;char map[102][102];int dir[8][2]={1,0,-1,0,0,1,0,-1,1,1,-1,1,1,-1,-1,-1};int n,m;void dfs(int x,int y){ for(int i=0;i<8;i++) { int dx=x+dir[i][0]; int dy=y+dir[i][1]; if(dx>=0&&dy>=0&&dx<n&&dy<m&&map[dx][dy]=='@') { map[dx][dy]='*'; dfs(dx,dy); } }}int main(){ int i,j; int sum; while(scanf("%d%d",&n,&m)!=EOF,n&&m) { sum=0; for(i=0;i<n;i++) scanf("%s",map[i]); for(i=0;i<n;i++) { for(j=0;j<m;j++) { if(map[i][j]=='@') { dfs(i,j); sum++; } } } printf("%d\n",sum); } return 0;}#include<stdio.h>#include<string.h>#include<algorithm>#include<stdlib.h>using namespace std;char map[102][102];int dir[8][2]={1,0,-1,0,0,1,0,-1,1,1,-1,1,1,-1,-1,-1};int n,m;void dfs(int x,int y){ for(int i=0;i<8;i++) { int dx=x+dir[i][0]; int dy=y+dir[i][1]; if(dx>=0&&dy>=0&&dx<n&&dy<m&&map[dx][dy]=='@') { map[dx][dy]='*'; dfs(dx,dy); } }}int main(){ int i,j; int sum; while(scanf("%d%d",&n,&m)!=EOF,n&&m) { sum=0; for(i=0;i<n;i++) scanf("%s",map[i]); for(i=0;i<n;i++) { for(j=0;j<m;j++) { if(map[i][j]=='@') { dfs(i,j); sum++; } } } printf("%d\n",sum); } return 0;}
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