pat 甲级 1091(bfs)

题目链接:https://pintia.cn/problem-sets/994805342720868352/problems/994805375457411072

思路:给出一个脑子的体积,0表示正常细胞,1表示中风细胞,求中风细胞的总体积

,如果连续的中风细胞的体积超过v,就记为中风细胞。

其实就是找一个连续的1细胞的体积,如果体积大于v就算,小于v就不算。用bfs就好了,开始我好傻,

还去用dfs。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int maxn1 = 60;
const int maxn2 = 130;
const int maxn3 = 1300;
int vis[maxn3][maxn2][maxn1],a[maxn3][maxn2][maxn1],n,m,k;
int xx[6]={0,0,1,0,0,-1};
int yy[6]={0,1,0,0,-1,0};
int zz[6]={1,0,0,-1,0,0};
struct Node{
	int x,y,z;
};
int bfs(int x,int y,int z)
{
	int i,j,ans=0;
	queue <Node> q;
	Node tmp,tp;
	tmp.x=x;tmp.y=y;tmp.z=z;
	q.push(tmp);
	while(!q.empty())
	{
		tmp=q.front();
		q.pop();
		ans++;
		for(i=0;i<6;i++)
		{
			tp.x=tmp.x+xx[i];
			tp.y=tmp.y+yy[i];
			tp.z=tmp.z+zz[i];
			if(tp.x<1||tp.y<1||tp.z<1||tp.x>n||tp.y>m||tp.z>k) continue;
			if(vis[tp.x][tp.y][tp.z]==0&&a[tp.x][tp.y][tp.z]==1)
			{
				vis[tp.x][tp.y][tp.z]=1;
				q.push(tp);
			}
		}
	}
	return ans;
}
int main(void)
{
	int i,j,q,l,v;
	scanf("%d%d%d%d",&m,&n,&k,&v);
	for(q=1;q<=k;q++)
	    for(j=1;j<=m;j++)
	        for(i=1;i<=n;i++)
	        scanf("%d",&a[i][j][q]);
	int ans=0;
	memset(vis,0,sizeof(vis));
	for(q=1;q<=k;q++)
	    for(j=1;j<=m;j++)
	        for(i=1;i<=n;i++)
	        {
	        	if(vis[i][j][q]==0&&a[i][j][q]==1)
	        	{
	        		vis[i][j][q]=1;
	        		int tp=bfs(i,j,q);
	        		if(tp>=v) ans+=tp;
				}
			}
	printf("%d\n",ans);
	return 0;
} 

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转载自blog.csdn.net/qq_41829060/article/details/84145987