VJ--图论---BFS搜索啊搜索，又改了好久，原因是：首先是第一个忘记标记啦，后来又发现队列开小啦，各种错啊！！

最新推荐文章于 2025-05-05 19:00:04 发布

原创最新推荐文章于 2025-05-05 19:00:04 发布 · 548 阅读

0 ·

CC 4.0 BY-SA版权

图论同时被 2 个专栏收录

3 篇文章

订阅专栏

BFS与DFS

2 篇文章

订阅专栏

本文介绍了一种用于探测地下油藏分布的算法。该算法通过分析网格中各单元格的状态来确定油藏的数量及其连通性。网格由星号(*)表示空地、@表示油藏单元。算法使用广度优先搜索(BFS)来遍历每个可能的油藏区域，并标记已访问过的油藏单元，最终统计出独立油藏的数量。

A - Oil Deposits

Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u

Submit Status

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

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char map[105][105];
int vis[105][105];
int mv[8][2]= {{1,0},{0,1},{-1,0},{0,-1},{1,1},{-1,1},{1,-1},{-1,-1}};
int s;
int m,n;
struct A
{
    int x;
    int y;
} p[10050];
int enter=0,out=0;
struct A f,t;
int getdian()
{
    int i,j;
    for(i=0; i<m; i++)
    {
        for(j=0; j<n; j++)
        {
            if(map[i][j]=='@'&&!vis[i][j])
                break;
        }
        if(j<n)
            break;
    }
    if(i>=m&&j>=n)
        return s;
    else
    {
        s++;
        vis[i][j]=1;
        BFS(i,j);
    }
}
void BFS(int x,int y)
{
    int i;
    f.x=x;
    f.y=y;
    p[enter++]=f;
    while(out<enter)
    {
        t=p[out++];
        for(i=0; i<8; i++)
        {
            f.x=t.x+mv[i][0];
            f.y=t.y+mv[i][1];
            if(0<=f.x&&f.x<m&&0<=f.y&&f.y<n&&!vis[f.x][f.y]&&map[f.x][f.y]=='@')
            {
                vis[f.x][f.y]=1;
                p[enter++]=f;
            }
        }
    }
    if(out>=enter)
        getdian();
}
int main()
{
    int i;
    int t;
    while(~scanf("%d%d",&m,&n)&&m&&n)
    {
        memset(vis,0,sizeof(vis));
        s=0;
        for(i=0; i<m; i++)
            scanf("%*c%s",map[i]);
        printf("%d\n",getdian());
    }
    return 0;
}