1091 Acute Stroke (30point(s)) 需要二刷 *典型的BFS问题，和五子棋问题或许有关联

基本思想：

三维BFS遍历，注意可以XYZ变换数组进行快速遍历；

 

关键点：

注意访问的标志位问题；

 

#include<iostream>
#include<stdlib.h>
#include<stdio.h>
#include<vector> 
#include<string>
#include<math.h>
#include<algorithm>
#include<cstring>
#include<map>
#include<queue>
#include<set>
#include<stack>
using namespace std;

int m, n, l, t;
int cnt = 0;

struct node {
    int x, y, z;
};

int matrix[1290][130][61];
bool flag[1290][130][61];

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 };

void init() {
    fill(matrix[0][0], matrix[0][0] + 1290 * 130 * 61, 0);
    fill(flag[0][0], flag[0][0] + 1290 * 130 * 61, true);
}

bool charge(int x, int y, int z) {
    if (x >= m || x<0 || y>=n || y<0 || z>=l || z < 0)
        return false;
    if (!flag[x][y][z]||matrix[x][y][z]==0)
        return false;
    return true;
}

int bfs(int x,int y,int z) {
    int ct = 0;
    queue<node>q;
    node no;
    no.x = x;
    no.y = y;
    no.z = z;
    q.push(no);
    while (!q.empty()){
        ct++;
        no = q.front();
        q.pop();
        flag[no.x][no.y][no.z] = false;
        //cout << no.x << " " << no.y << " " << no.z << endl;
        for (int i = 0; i < 6; i++) {
            int a = no.x + X[i];
            int b = no.y + Y[i];
            int c = no.z + Z[i];
            if (charge(a,b,c)) {
                node noo;
                noo.x = a;
                noo.y = b;
                noo.z = c;
                q.push(noo);
                flag[a][b][c] = false;
            }
        }
    }
    if (ct >= t)
        return ct;
    else
        return 0;
}

int main() {
    init();
    cin >> m >> n >> l >> t;
    for (int z = 0; z < l; z++) {
        for (int x = 0; x < m; x++) {
            for (int y = 0; y < n; y++) {
                scanf("%d", &matrix[x][y][z]);
            }
        }
    }
    for (int z = 0; z < l; z++) {
        for (int x = 0; x < m; x++) {
            for (int y = 0; y < n; y++) {
                if (flag[x][y][z]&&matrix[x][y][z]==1)
                    cnt += bfs(x, y, z);
            }
        }
    }
    cout << cnt << endl;
    return 0;
}