1004 Acute Stroke

题目描述
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.

输入描述:
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.

输出描述:
For each case, output in a line the total volume of the stroke core.

输入例子:
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

输出例子:
26

package p1004;import java.util.LinkedList;import java.util.Queue;import java.util.Scanner;public class Main {    static class Point {        int x, y, z;        public Point(int x, int y, int z) {            super();            this.x = x;            this.y = y;            this.z = z;        }    }    public static void main(String[] args) {        Scanner sc = new Scanner(System.in);        int m = sc.nextInt();        int n = sc.nextInt();        int l = sc.nextInt();        int threshold = sc.nextInt();        int[][][] data = new int[m][n][l];        for(int i=0; i<l; i++)            for(int j=0; j<m; j++)                for(int k=0; k<n; k++)                    data[j][k][i] = sc.nextInt();        Queue<Point> q = new LinkedList<Point>();        boolean[][][] visited = new boolean[m][n][l];        int[] dx = new int[]{1,-1,0,0,0,0};        int[] dy = new int[]{0,0,1,-1,0,0};        int[] dz = new int[]{0,0,0,0,1,-1};        int rst = 0, sum = 1;        for(int i=0; i<l; i++)            for(int j=0; j<m; j++)                for(int k=0; k<n; k++) {                    if(visited[j][k][i] == false && data[j][k][i] == 1) {                        visited[j][k][i] = true;                        q.clear();                        q.add(new Point(j, k, i));                        sum = 1;                        while(!q.isEmpty()) {                            Point p = q.remove();                            int x = p.x, y = p.y, z = p.z;                            for(int cnt=0; cnt<6; cnt++) {                                int xx = x+dx[cnt], yy = y+dy[cnt], zz = z+dz[cnt];                                if((xx>=0&&xx<m) && (yy>=0&&yy<n) && (zz>=0&&zz<l) && visited[xx][yy][zz] == false && data[xx][yy][zz] == 1) {                                    q.add(new Point(xx, yy, zz));                                    visited[xx][yy][zz] = true;                                    sum ++;                                }                            }                        }                        if(sum >= threshold)                            rst += sum;                    }                }        System.out.println(rst);    }}

只能说自己的英语还是太挫了，表示没看懂题目。。。。

思路就是三维BFS，有一个技巧是设置三个数组简化代码

int[] dx = new int[]{1,-1,0,0,0,0};int[] dy = new int[]{0,0,1,-1,0,0};int[] dz = new int[]{0,0,0,0,1,-1};