标签：queue center nts ret slice 左右 ase sample rate
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 by 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 by 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.
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 0Sample Output:
26
思路
由题意，M行N列的矩阵表示一片脑部切图，矩阵中的值——0表示该位置一切正常，1表示该位置为肿瘤区，L表示脑部切图的片数，由此构建了一个M、N、L的三维的矩阵。
现在让你确定统计肿瘤区的个数，注意只有上下前后左右相邻的肿瘤区且这些相邻的区域个数必须不小于T才能被统计。
归根结底其实就是一个3D图的连通分量问题，即统计每一个节点数不小于T的连通分量的节点数之和。
用BFS或者DFS都行，这里用BFS更好，DFS递归可能会栈溢出。
代码
#include<iostream>
#include<queue>
using namespace std;
/*
坐标系如下
xxxxxxxxxxxxxxxxxx + →
z
z
z
z
z
z
z
+
↓
y轴垂直屏幕射出
*/
class node
{
public:
int x,z,y;
node(int a,int b, int c){x = a;z = b;y = c;}
};
//6个方向分别为上下前后左右
int X[6] = {0,0,0,0,-1,1};
int Y[6] = {1,-1,0,0,0,0};
int Z[6] = {0,0,1,-1,0,0};
int graph[1290][130][65];
bool visit[1290][130][65];
int M,N,L,T; // z x y T
bool check(int x,int z,int y)
{
if(x < 0 || z < 0 || y < 0 || x >= N || z >= M || y >= L)
return false;
if(graph[x][z][y] == 0 || visit[x][z][y])
return false;
return true;
}
int bfs(int x,int z,int y)
{
int cnt = 0;
node tmp(x,z,y);
queue<node> q;
q.push(tmp);
visit[x][z][y] = true;
while(!q.empty())
{
node f = q.front();
q.pop();
cnt++;
for(int i = 0;i < 6;i++)
{
int newx = f.x + X[i];
int newz = f.z + Z[i];
int newy = f.y + Y[i];
if(check(newx,newz,newy))
{
visit[newx][newz][newy] = true;
tmp.x = newx;
tmp.y = newy;
tmp.z = newz;
q.push(tmp);
}
}
}
if(cnt >= T)
return cnt;
else
return 0;
}
int main()
{
cin >> M >> N >> L >> T;
for(int y = 0;y < L;y++)
{
for(int z = 0; z < M;z++)
{
for(int x = 0;x < N;x++)
{
cin >> graph[x][z][y];
visit[x][z][y] = false;
}
}
}
int res = 0;
for(int y = 0;y < L;y++)
{
for(int z = 0; z < M;z++)
{
for(int x = 0;x < N;x++)
{
if(graph[x][z][y] == 1 && !visit[x][z][y])
res += bfs(x,z,y);
}
}
}
cout << res << endl;
}
标签：queue center nts ret slice 左右 ase sample rate
原文地址：http://www.cnblogs.com/0kk470/p/7881443.html
