HDU1241 Oil Deposits

                                                               Oil Deposits

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7709    Accepted Submission(s): 4514

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 Input

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

 

Sample Output

0122

 

SourceMid-Central USA 1997
 

RecommendEddy

解題思路：廣度優先搜尋，圖的連通分量。

遍曆圖——>找到油田——>把油田群找出來——>標記、計數——>繼續遍曆——>遍曆完成

只要相鄰既是一起，也就是說，遍曆方向有8個，所有相鄰的都算。

#include<cstdio>#include<cstring>#include<queue>using namespace std;int n,m;char map[100][100];int sum;int dir[8][2]={1,0,1,1,1,-1,0,1,0,-1,-1,0,-1,1,-1,-1};      //8個方向struct node{    int x;    int y;};int inmap(int x,int y){    if(x>=0&&x<n&&y>=0&&y<m)        return true;    return false;}void bfs(int x,int y){    int i;    node u,v;    queue<node> q;    u.x=x;    u.y=y;    q.push(u);    while(!q.empty())    {        u=q.front();        q.pop();        for(i=0;i<8;i++)        {            v.x=u.x+dir[i][0];            v.y=u.y+dir[i][1];            if(inmap(v.x,v.y)&&map[v.x][v.y]=='@')     //是油田哦            {                map[v.x][v.y]='*';     //油田群遍曆，標記                q.push(v);            }        }    }}int main(){    int i,j;    while(scanf("%d%d",&n,&m)&&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]=='@')      //找到油田                {                    sum++;                    bfs(i,j);                }            }        }        printf("%d\n",sum);    }    return 0;}