PAT Acute Stroke (30)

題目描述

One important factor to identify acute stroke (急性腦卒中) 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.

 

輸入描述:

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.

Figure 1

輸出描述:

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

 

輸入例子:

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

 

輸出例子:

26


題目描述的不清晰，看了好半天都沒看懂😢
大意是一個M*N*L的點陣，三維點陣，每一個點取值0或者1，每一個取值爲1的點和最近的6個點中取值爲1的點連通（上下先後左右）
忽略1的個數小於T的連通子圖，而後把其他的連通子圖的1的個數相加，最後輸出相加結果。
因此用廣度優先就能夠解決了。

//
//  Acute Stroke (30).cpp
//  PAT_test
//
//  Created by 彭威 on 15/8/6.
//  Copyright © 2015年 biophy.nju.edu.cn. All rights reserved.
//

#include <iostream>
#include <vector>
#include <queue>
using namespace std;
vector<vector<vector<int>>> Q;
vector<vector<vector<bool>>> V;
struct xyz {
    int a[3];
    xyz(int i,int j,int k){
        a[0]=i;
        a[1]=j;
        a[2]=k;
    }
    bool valid(){
        return (!V[a[0]][a[1]][a[2]] && Q[a[0]][a[1]][a[2]]);
    }
};
vector<xyz> validNeighbor(xyz p){
    xyz temp(p);
    vector<xyz> re;
    for (int i=0; i<3; i++) {
        temp.a[i]+=1;
        if (temp.valid()) {
            re.push_back(temp);
        }
        temp.a[i]-=2;
        if (temp.valid()) {
            re.push_back(temp);
        }
        temp.a[i]+=1;
    }
    
    return re;
}
int BFS(int i,int j,int k){
    queue<xyz> q;
    int m=0;
    xyz p(i, j, k);
    V[p.a[0]][p.a[1]][p.a[2]]=true;
    q.push(p);
    while (!q.empty()) {
        xyz p = q.front();
        q.pop();
        ++m;
        vector<xyz> x=validNeighbor(p);
        for (int i=0; i<x.size(); i++) {
            V[x[i].a[0]][x[i].a[1]][x[i].a[2]]=true;
            q.push(x[i]);
        }
    }
    return m;
}
int main(int argc,const char* argv[]){
    ios::sync_with_stdio(false);
    int M,N,L,T;
    long long VV=0;
    cin>>M>>N>>L>>T;
    Q.resize(L+2);
    V.resize(L+2);
    for (int i=0; i<L+2; i++) {
        Q[i].resize(M+2);
        V[i].resize(M+2);
        for (int j=0; j<M+2; j++) {
            Q[i][j].resize(N+2, 0);
            V[i][j].resize(N+2, false);
        }
    }
    for (int i=1; i<L+1; i++) {
        for (int j=1; j<M+1; j++) {
            for (int k=1; k<N+1; k++) {
                cin>>Q[i][j][k];
            }
        }
    }
    for (int i=1; i<L+1; i++) {
        for (int j=1; j<M+1; j++) {
            for (int k=1; k<N+1; k++) {
                if(Q[i][j][k] && !V[i][j][k]){
                    int m = BFS(i,j,k);
                    if (m<T) {
                        continue;
                    }
                    VV+=m;
                }
            }
        }
    }
    cout<<VV<<endl;
    return 0;
}