[转]PAT甲级练习1091. Acute Stroke (30)

1091. Acute Stroke (30)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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.

Figure 1

Output Specification:

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

Sample Input:

3 4 5 21 1 1 11 1 1 11 1 1 10 0 1 10 0 1 10 0 1 11 0 1 10 1 0 00 0 0 01 0 1 10 0 0 00 0 0 00 0 0 10 0 0 11 0 0 0

Sample Output:

26

大意是求连通域的大小，不同于平面，是一个三维空间上的

http://www.liuchuo.net/archives/2307 http://blog.csdn.net/sinat_29278271/article/details/47768209

题目大意：给定一个三维数组，0表示正常1表示有肿瘤，肿瘤块的大小大于等于t才算作是肿瘤，让计算所有满足肿瘤块的大小
分析：三维的广度优先搜索~~XYZ三个数组判断方向，对每一个点广度优先累计肿瘤块的大小，如果大于等于t就把结果累加。用visit数组标记当前的点有没有被访问过，被访问过的结点是不会再访问的。。judge判断是否超过了边界，或者是否当前结点为0不是肿瘤~~~

#include <cstdio>#include <queue>using namespace std;struct node {    int x, y, z;};int m, n, l, t;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};int arr[1300][130][80];bool visit[1300][130][80];bool judge(int x, int y, int z) {    if(x < 0 || x >= m || y < 0 || y >= n || z < 0 || z >= l) return false;    if(arr[x][y][z] == 0 || visit[x][y][z] == true) return false;    return true;}int bfs(int x, int y, int z) {    int cnt = 0;    node temp;    temp.x = x, temp.y = y, temp.z = z;    queue<node> q;    q.push(temp);    visit[x][y][z] = true;    while(!q.empty()) {        node top = q.front();        q.pop();        cnt++;        for(int i = 0; i < 6; i++) {            int tx = top.x + X[i];            int ty = top.y + Y[i];            int tz = top.z + Z[i];            if(judge(tx, ty, tz)) {                visit[tx][ty][tz] = true;                temp.x = tx, temp.y = ty, temp.z = tz;                q.push(temp);            }        }    }    if(cnt >= t)        return cnt;    else        return 0;    }int main() {    scanf("%d %d %d %d", &m, &n, &l, &t);    for(int i = 0; i < l; i++)        for(int j = 0; j < m; j++)            for(int k = 0; k < n; k++)                scanf("%d", &arr[j][k][i]);    int ans = 0;    for(int i = 0; i < l; i++) {        for(int j = 0; j < m; j++) {            for(int k = 0; k < n; k++) {                if(arr[j][k][i] == 1 && visit[j][k][i] == false)                    ans += bfs(j, k, i);            }        }    }    printf("%d", ans);    return 0;}