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
note
- 题意
a. 第一遍理解错误以为是slice个二位数组，考到图有感到奇怪，但是没有想到slice是第三维
b. 题意：三维数组，相邻1的个数大于阈值则计数，求总数 - 思路
a. 进行slice遍查找，每次dfs一边整个二维数组
b. 利用三位数组存储，a和visited均是三位，bfs，原因不涉及递归解决内存问题
Code1
include<iostream>
using namespace std;
const int maxn = 1e4;
int more_than_threshold = 0, sum = 0, visited_num;
int step = 0, length, width, slice, threshold;
int a[maxn][maxn];
int visited[maxn][maxn];
void Dfs(int x, int y){
if(visited[x][y] == 1) return;
visited_num++;
if(a[x][y] == 0) return;
visited[x][y] = 1;
step++;// 一次dfs搜索到的1的个数
for(int i = -1; i < 2; i++){
for(int j = -1; j < 2; j++){
if(visited[x + i][y + j] == 0 && (i != 0 || j != 0)){
Dfs(x + i, y + j);
}
}
}
}
int main(){
cin >> length >> width >> slice >> threshold;
for(int time = 0; time < slice; time++){
fill(a[0], a[0] + (length + 1) * (length + 1), 0);
fill(visited[0], visited[0] + (maxn + 1) * (maxn + 1), 0);
visited_num = 0;
for(int i = 1; i <= length; i++){
for(int j = 1; j <= width; j++){
cin >> a[i][j];
}
}
for(int i = 1; i <= length; i++){
for(int j = 1; j <= width; j++){
step = 0;
Dfs(i, j);
if(visited_num >= length*width) break;
if(step >= threshold) sum += step;
}
}
}
cout << sum;
return 0;
}