1091 Acute Stroke (30 分)

1091 Acute Stroke (30 分)
One important factor to identify acute stroke (急性脑卒中) is the volume of the stroke core. Given the results of image analysis in which the core regions are identified in each MRI slice, your job is to calculate the volume of the stroke core.

Input Specification:
Each input file contains one test case. For each case, the first line contains 4 positive integers: M, N, L and T, where M and N are the sizes of each slice (i.e. pixels of a slice are in an M×N matrix, and the maximum resolution is 1286 by 128); L (≤60) is the number of slices of a brain; and T is the integer threshold (i.e. if the volume of a connected core is less than T, then that core must not be counted).

Then L slices are given. Each slice is represented by an M×N matrix of 0’s and 1’s, where 1 represents a pixel of stroke, and 0 means normal. Since the thickness of a slice is a constant, we only have to count the number of 1’s to obtain the volume. However, there might be several separated core regions in a brain, and only those with their volumes no less than T are counted. Two pixels are connected and hence belong to the same region if they share a common side, as shown by Figure 1 where all the 6 red pixels are connected to the blue one.

Figure 1

Output Specification:
For each case, output in a line the total volume of the stroke core.

Sample Input:
3 4 5 2
1 1 1 1
1 1 1 1
1 1 1 1
0 0 1 1
0 0 1 1
0 0 1 1
1 0 1 1
0 1 0 0
0 0 0 0
1 0 1 1
0 0 0 0
0 0 0 0
0 0 0 1
0 0 0 1
1 0 0 0
Sample Output:
26

题目大意：给定一个三维数组，数组元素为1时代表有肿块，数组元素为0时表示没有肿块，肿块的大小大于等于给定的数值时才能加入总体积
分析：设置X/Y/Z数组判断方向，每次遍历这三个数组时相当于遍历与一个像素点相邻的六个像素点。定义panduan函数判断该像素点是否可以加入。设置visit标记结点有没有被访问，每个结点只能被访问一次。用广度优先搜索的模板实现

AC代码

#include<bits/stdc++.h>
using namespace std;
int X[6] = {
    
    1, 0, 0, -1, 0, 0};
int Y[6] = {
    
    0, 1, 0, 0, -1, 0};
int Z[6] = {
    
    0, 0, 1, 0, 0, -1};//记住三维的构造方式
int region[1300][130][70];
bool visit[1300][130][70];
int M, N, L, T,sum=0;
struct node{
    
    
  int x, y, z;
};
int panduan(int x,int y,int z){
    
    
  if(x<=0||x>N||y<=0||y>M||z<=0||z>L)
    return false;
  if(visit[x][y][z]==true||region[x][y][z]==0)
    return false;

  return true;
}
/*熟记广度优先搜索的模板
1.任选图中一个结点访问，入队，并将这个顶点标记为已访问
2.当队列不空时循环执行：出队，依次检查出队顶点的所有邻接顶点，访问没有被访问过的邻接顶点并将其入队
3.当队列为空时跳出循环，广度优先搜索完成
*/
int BFS(node topNode){
    
    
  int cnt = 0;
  queue<node> q;
  visit[topNode.x][topNode.y][topNode.z] = true;//将入队结点标记为已访问
  q.push(topNode);
  int x, y, z;
  while(!q.empty()){
    
    
    node temp = q.front();
    q.pop();
    cnt++;//每一个结点都需要被抛出，记录抛出结点的个数
    for (int i = 0; i < 6;i++){
    
    
      x=temp.x + X[i];
      y=temp.y + Y[i];
      z=temp.z + Z[i];
      if(panduan(x,y,z)){
    
    
        topNode = {
    
    x, y, z};
        q.push(topNode);
        visit[x][y][z] = true;
      }//检查出队顶点的所有邻接顶点
    }
  }
  if(cnt>=T)
  return cnt;
  else
    return 0;
}

int main(){
    
    
  cin >> M >> N >> L >> T;
  for (int i = 1; i <= L;i++){
    
    
    for (int j = 1; j <= M;j++){
    
    
      for (int k = 1; k <= N;k++)
        cin >> region[k][j][i];
    }
  }
  for (int i = 1; i <= L;i++){
    
    
    for (int j = 1; j <= M;j++){
    
    
      for (int k = 1; k <= N;k++){
    
    
        if(visit[k][j][i]==false&&region[k][j][i]==1){
    
    
        node temp = {
    
    k, j, i};
        sum += BFS(temp);
        }
      } 
    }
  }
  printf("%d", sum);

  return 0;
}