PAT Acute Stroke (30)

题目描述

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.

输入描述:

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

输出描述:

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

输入例子:

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

输出例子:

26

题目描述的不清晰，看了好半天都没看懂??大意是一个M*N*L的点阵，三维点阵，每个点取值0或者1，每个取值为1的点和最近的6个点中取值为1的点连通（上下前后左右）忽略1的个数小于T的连通子图，然后把其余的连通子图的1的个数相加，最后输出相加结果。所以用广度优先就可以解决了。

//
//  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;
}