1091. Acute Stroke(30)

看到这道最先想到的是数连通集，个人喜好，利用递归的方式进行深度遍历，可惜最后两个用例 段错误

1.深度遍历.

代码如下：

#include<iostream>
#include<cstdio>
#include<vector>
using namespace std;
const int maxm = 1300;
const int maxn = 130;
const int maxl = 62;
const int xc[] = {-1,0,1,0,0,0};
const int yc[] = {0,-1,0,1,0,0};
const int zc[] = {0,0,0,0,1,-1};
typedef struct Piece
{
	int p[maxm][maxn];
}Piece;
vector<Piece> pic;
int M,N,L,T;
int total;
int cur;

int inbound(int x,int y,int z)
{
	if(x>=0 && x<L &&y>=0 &&y<M && z>=0 && z<N)
		return 1;
	return 0;
}

void calcu(int i,int j,int k)
{
	cur += pic[i].p[j][k];
	pic[i].p[j][k] = 0;
//printf("cur = %d,visited[%d][%d][%d]\n",cur,i,j,k);
	int u;
	for(u = 0;u<6;++u)
	{
		int x = i+xc[u];
		int y = j+yc[u];
		int z = k+zc[u];
		if(inbound(x,y,z) && pic[x].p[y][z])
		{
			calcu(x,y,z);
		}
	}
}

void build()
{
	int i,j,k;
	for(i = 0; i<L;++i)
	{
		for(j = 0; j<M;++j)
		{
			for(k = 0;k<N;++k)
			{
				if(pic[i].p[j][k] == 1)
				{
					cur = 0;
					calcu(i,j,k);
					if(cur>=T)
					{
						total += cur;
					}
				}
			}
		}
	}	
}


int main()
{
freopen("0.txt","r",stdin);
	cin>>M>>N>>L>>T;
	int i,j,k;
	for(i = 0; i<L;++i)
	{
		pic.push_back(Piece());
		Piece & pc = pic[pic.size()-1];
		for(j = 0; j<M;++j)
		{
			for(k = 0; k<N;++k)
			{
				cin>>pc.p[j][k];
			}
		}
	}
	total = 0;
	build();
	cout<<total<<"\n";
	return 0;
}

很可惜，利用深度优先，只能拿到25分，会发生段错误，还以为是存储花了太多内存

后来三个维度都用vector来存储，还是最后两个用例是段错误。

改用队列记录的广度优先遍历：

//有个博客说并查集，很复杂，看了第二个博客，用广度优先遍历
//利用广度优先遍历，不再有段错误，看来深度遍历的栈，还是比广度遍历自己维护的队列要费空间的
//这个代码是满分
#include<iostream>
#include<cstdio>
#include<vector>
#include<queue>
using namespace std;
const int maxm = 1300;
const int maxn = 130;
const int maxl = 62;
const int xc[] = {-1,0,1,0,0,0};
const int yc[] = {0,-1,0,1,0,0};
const int zc[] = {0,0,0,0,1,-1};
typedef vector<int> Row;
typedef vector<Row> Piece;
vector<Piece> pic;
int M,N,L,T;
int total;
int cur;
typedef struct Loc
{
	int x,y,z;
	Loc(){}
	Loc(int x,int y, int z):x(x),y(y),z(z){}
}Loc;

int inbound(int x,int y,int z)
{
	if(x>=0 && x<L &&y>=0 &&y<M && z>=0 && z<N)
		return 1;
	return 0;
}

void calcu(int i,int j,int k)
{
	queue<Loc> q;
	q.push(Loc(i,j,k));
	pic[i][j][k] = 0;
	++cur;
	while(!q.empty())
	{
		Loc l = q.front();
		q.pop();
		int u;
		for(u = 0;u<6;++u)
		{
			int x = l.x+xc[u];
			int y = l.y+yc[u];
			int z = l.z+zc[u];
			if(inbound(x,y,z) && pic[x][y][z])
			{
				q.push(Loc(x,y,z));
				pic[x][y][z] = 0;
				++cur;
			}
		}
	}
}

void build()
{
	int i,j,k;
	for(i = 0; i<L;++i)
	{	
		for(j = 0; j<M;++j)
		{
			for(k = 0;k<N;++k)
			{
				if(pic[i][j][k] == 1)
				{
					cur = 0;
					calcu(i,j,k);
					if(cur>=T)
					{
						total += cur;
					}
				}
			}
		}
	}	
}


int main()
{
//freopen("0.txt","r",stdin);
	cin>>M>>N>>L>>T;
	int i,j,k;
	for(i = 0; i<L;++i)
	{
		pic.push_back(Piece());
		Piece &p = pic[pic.size()-1];
		for(j = 0; j<M;++j)
		{
			p.push_back(Row());
			Row &r = p[p.size()-1];
			for(k = 0; k<N;++k)
			{
				int buf;
				cin>>buf;
				r.push_back(buf);
			}
		}
	}
	total = 0;
	build();
	cout<<total<<"\n";
	return 0;
}

最后两个用例的分才拿下。