poj1562 Oil Deposits BFS

Oil Deposits

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 12620 Accepted: 6887

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 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

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

Source

Mid-Central USA 1997

#include<stdio.h>#include<stdlib.h>int fx[8][2]= {{0,1},{0,-1},{1,0},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1}};//八个方向char map[101][101];//储存地图int m,n;void sou(int x,int y){    int i,tx,ty;    for(i=0; i<8; i++)//八个方向搜索    {        tx=x+fx[i][0];//搜索        ty=y+fx[i][1];        if(tx<0||tx>=m||ty<0||ty>=n)//越界           continue;        if(map[tx][ty]=='*')//陆地            continue;        if(map[tx][ty]=='@')//油田        {            map[tx][ty]='*';//油田变陆地，防止回溯循环        }        sou(tx,ty);    }}int main(){    int i,j,count;    while(scanf("%d%d",&m,&n),m!=0||n!=0)    {        for(i=0; i<m; i++)            scanf("%s",map[i]);        count=0;        for(i=0; i<m; i++)            for(j=0; j<n; j++)            {                if(map[i][j]=='@')                {                    sou(i,j);//无路可搜，油田+1                    count++;                }            }        printf("%d\n",count);    }    return 0;}

<strong><span style="font-size: 18px; color: rgb(255, 0, 0);">搜索到油田将之变为陆地（防止回溯无限搜索），然后继续向八个方向搜索油田，进而变为陆地，直到搜不到油田为止，油田数目加一；然后进行下一次搜索。</span></strong>