PAT(Advanced Level)A1091. Acute Stroke

题意

其实就是一个三维BFS的问题，计算满足大小的块（>=T）一共有几个

思路

• 其实就是二维BFS扩展到三维空间的情况，无非就是把本来二维上扩展4个方向改成扩展6个方向

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

• 统计入队元素，看有没有>=T即可。几乎就是裸模板题了～

代码

#include <algorithm>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
const int dx[] = {0, -1, 0, 1, 0, 0};
const int dy[] = {-1, 0, 1, 0, 0, 0};
const int dz[] = {0, 0, 0, 0, 1, -1};
struct Node{
    int x, y, z;
};

int cub[1300][150][80] = {0};
bool vis[1300][150][80] = {false};
int M, N, L, T;
int level = 0;
int sum = 0;
bool judge(int x, int y, int z)
{
    if(x < 0 || x >= M ||
       y < 0 || y >= N ||
       z < 0 || z >= level)
        return false;
    if(cub[x][y][z] == 0 || vis[x][y][z] == true)   return false;
    return true;
}
void bfs(int x, int y, int z)
{
    int cur_sum = 1;
    Node cur;
    cur.x = x; cur.y = y; cur.z = z;
    queue<Node> q;
    q.push(cur);
    vis[x][y][z] = true;
    while(!q.empty())
    {
        auto tmp = q.front();
        q.pop();
        for(int i=0;i<6;i++)
        {
            int tx = tmp.x + dx[i];
            int ty = tmp.y + dy[i];
            int tz = tmp.z + dz[i];
            if(judge(tx, ty, tz))
            {
                vis[tx][ty][tz] = true;
                cur.x = tx, cur.y = ty, cur.z = tz;
                q.push(cur);
                cur_sum++;
            }
        }
    }
    if(cur_sum >= T)
        sum += cur_sum;
}

int main()
{
    cin >> M >> N >> L >> T;
    while(L--)
    {
        for(int i=0;i<M;i++)
            for(int j=0;j<N;j++)
                scanf("%d", &cub[i][j][level]);
        level++;
    }
    for(int k=0;k<level;k++)
        for(int i=0;i<M;i++)
            for(int j=0;j<N;j++)
                    if(cub[i][j][k] == 1 && !vis[i][j][k])
                        bfs(i, j, k);
    cout << sum << endl;
    return 0;
}