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UVA Oil Deposits (BFS)

 Oil Deposits 

UVA       Oil Deposits (BFS)_ios


Sample Input 

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


Sample Output 



0
1
2
2



       题意:问地图中有多少个@,前提如果@的八个方向中如果还是@,则相邻的@不进入计数。(简单的BFS)



代码:




#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<string>
#include<queue>

using namespace std;

struct node
{
    int x;
    int y;
} q[10010];
int jx[] = {0,1,0,-1,1,1,-1,-1};
int jy[] = {1,0,-1,0,1,-1,1,-1};
int n,m;
char map[101][101];
int v[101][101];
int pt;
void BFS()
{
    node t,f;
    memset(v,0,sizeof(v));
    queue<node>p;
    int count = 0;
    for(int i=0; i<pt; i++)
    {
        if(v[q[i].x][q[i].y] == 0)
        {
            t.x = q[i].x;
            t.y = q[i].y;
            v[t.x][t.y] = 1;
            count++;
            p.push(t);
            while(!p.empty())
            {
                t = p.front();
                p.pop();
                for(int j=0;j<8;j++)
                {
                    f.x = t.x + jx[j];
                    f.y = t.y + jy[j];
                    if(f.x>=0 && f.x<n && f.y>=0 && f.y<m && map[f.x][f.y] == '@' && v[f.x][f.y] == 0)
                    {
                        p.push(f);
                        v[f.x][f.y] = 1;
                    }
                }
            }
        }
    }
    printf("%d\n",count);

}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        if(n == 0 && m == 0)
        {
            break;
        }
        pt = 0;
        struct node a;
        for(int i=0; i<n; i++)
        {
            scanf("%s",map[i]);
            for(int j=0; j<m; j++)
            {
                if(map[i][j] == '@')
                {
                    a.x = i;
                    a.y = j;
                    q[pt++] = a;
                }
            }
        }
        BFS();
    }
    return 0;
}



Miguel A. Revilla

1998-03-10




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