PAT甲级——Acute Stroke

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

思路：枚举三维数组中的每一个位置，如果为0则跳过；如果为1，则使用BFS查询与该位置相邻的6个位置（前提是不出界），判断它们是否为1（如果某个相邻的位置为1，则同样去查询与该位置相邻的6个位置，直到整个“1”块访问完毕）。

代码:

#include <iostream>
#include<queue>
using namespace std;struct node {int x, y, z;
}Node;int n, m, slice, T;//矩阵为n*m，共有slice层，T为卒中核心区中1的个数的下限
int pixel[1290][130][61];
bool inq[1290][130][61] = { false };
int X[6] = { 0,0,0,0,1,-1 };
int Y[6] = { 0,0,1,-1,0,0 };
int Z[6] = { 1,-1,0,0,0,0 };bool judge(int x, int y, int z)
{//越界返回falseif (x >= n || x < 0 || y >= m || y < 0 || z >= slice || z < 0) return false;//若当前位置为0或者（x,y,z)已经入过队，则返回falseif (pixel[x][y][z] == 0 || inq[x][y][z] == true) return false;//以上都不满足，返回truereturn true;
}int BFS(int x, int y, int z)
{int tot = 0;//计数当前块中1的个数queue<node>Q;Node.x = x, Node.y = y, Node.z = z;Q.push(Node);inq[x][y][z] = true;while (!Q.empty()){node top = Q.front();Q.pop();tot++;for (int i = 0; i < 6; i++){int newX = top.x + X[i];int newY = top.y + Y[i];int newZ = top.z + Z[i];if (judge(newX, newY, newZ)) {Node.x = newX, Node.y = newY, Node.z = newZ;Q.push(Node);inq[newX][newY][newZ] = true;}}}if (tot >= T) return tot;else return 0;
}
int main()
{scanf("%d%d%d%d", &n, &m, &slice, &T);for (int z = 0; z < slice; z++){for (int x = 0; x < n; x++){for (int y = 0; y < m; y++){scanf("%d", &pixel[x][y][z]);}}}int ans = 0;for (int z = 0; z < slice; z++){for (int x = 0; x < n; x++){for (int y = 0; y < m; y++) {if (pixel[x][y][z] == 1 && inq[x][y][z] == false) {ans += BFS(x, y, z);}}}}printf("%d\n", ans);return 0;
}