PAT(A) - 1091. Acute Stroke (30)

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

思路分析：第一次做这种三维的BFS，虽然思路是一样的，但还是没写出来.....英语也不好理解。参考了网上别人的题解。

#include <cstdio>#include <queue>using namespace std;int MGraph[1290][130][61];int visit[1290][130][61];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 };int m, n, l, T;typedef struct Point {    Point( int xx, int yy, int zz ) {        x = xx;        y = yy;        z = zz;    }    int x;    int y;    int z;} Point;bool judge( Point p ) {    // 越界    if( p.x >= m || p.x < 0 || p.y >= n || p.y < 0 || p.z >= l || p.z < 0 ) return false;    // 0的区域不可访问    if( MGraph[p.x][p.y][p.z] == 0 ) return false;    // 已经访问过    if( visit[p.x][p.y][p.z] ) return false;    // 其他情况返回true    return true;}int bfs( Point start ) {    queue<Point> q;    q.push( start );    visit[start.x][start.y][start.z] = 1;    int tot = 0;    while( !q.empty() ) {        Point cur = q.front();        q.pop();        tot++;        for( int i = 0; i < 6; i++ ) {  // 枚举6个方向            int newX = cur.x + X[i];            int newY = cur.y + Y[i];            int newZ = cur.z + Z[i];            Point p( newX, newY, newZ );            if( judge( p ) ) {                q.push( p );                visit[p.x][p.y][p.z] = 1;            }        }    }    return tot;}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", &MGraph[j][k][i] );            }    int sum = 0;    for( int i = 0; i < l; i++ )        for( int j = 0; j < m; j++ )            for( int k = 0; k < n; k++ ) {                if( !visit[j][k][i] && MGraph[j][k][i] == 1 ) {                    Point p( j, k, i );                    int ans = bfs( p );                    if( ans >= T ) {    // 大于等于阈值累计和                        sum += ans;                    }                }            }    printf( "%d\n", sum );    return 0;}