VK Cup 2016 - Qualification Round 2 D. Three-dimensional Turtle Super Computer 暴力

D. Three-dimensional Turtle Super Computer

题目连接：

http://www.codeforces.com/contest/638/problem/D

Description

A super computer has been built in the Turtle Academy of Sciences. The computer consists of n·m·k CPUs. The architecture was the paralellepiped of size n × m × k, split into 1 × 1 × 1 cells, each cell contains exactly one CPU. Thus, each CPU can be simultaneously identified as a group of three numbers from the layer number from 1 to n, the line number from 1 to m and the column number from 1 to k.

In the process of the Super Computer's work the CPUs can send each other messages by the famous turtle scheme: CPU (x, y, z) can send messages to CPUs (x + 1, y, z), (x, y + 1, z) and (x, y, z + 1) (of course, if they exist), there is no feedback, that is, CPUs (x + 1, y, z), (x, y + 1, z) and (x, y, z + 1) cannot send messages to CPU (x, y, z).

Over time some CPUs broke down and stopped working. Such CPUs cannot send messages, receive messages or serve as intermediates in transmitting messages. We will say that CPU (a, b, c) controls CPU (d, e, f) , if there is a chain of CPUs (xi, yi, zi), such that (x1 = a, y1 = b, z1 = c), (xp = d, yp = e, zp = f) (here and below p is the length of the chain) and the CPU in the chain with number i (i < p) can send messages to CPU i + 1.

Turtles are quite concerned about the denial-proofness of the system of communication between the remaining CPUs. For that they want to know the number of critical CPUs. A CPU (x, y, z) is critical, if turning it off will disrupt some control, that is, if there are two distinctive from (x, y, z) CPUs: (a, b, c) and (d, e, f), such that (a, b, c) controls (d, e, f) before (x, y, z) is turned off and stopped controlling it after the turning off.

Input

The first line contains three integers n, m and k (1 ≤ n, m, k ≤ 100) — the dimensions of the Super Computer.

Then n blocks follow, describing the current state of the processes. The blocks correspond to the layers of the Super Computer in the order from 1 to n. Each block consists of m lines, k characters in each — the description of a layer in the format of an m × k table. Thus, the state of the CPU (x, y, z) is corresponded to the z-th character of the y-th line of the block number x. Character "1" corresponds to a working CPU and character "0" corresponds to a malfunctioning one. The blocks are separated by exactly one empty line.

Output

Print a single integer — the number of critical CPUs, that is, such that turning only this CPU off will disrupt some control.

Sample Input

2 2 3
000
000

111
111

Sample Output

2

Hint

题意

有一个三维空间，从(x,y,z)的信号可以传输到(x+1,y,z),(x,y+1,z),(x,y,z+1)

现在有一些是坏的，就是不能用的

然后问你里面有多少个关键点。

关键点就是，如果这个坏了，会改变原来的传输信号的过程。

假设原来(x,y,z)能够传输到(x1,y1,z1)，但是由于坏了(x2,y2,z2)，导致不能传输了，就说(x2,y2,z2)是关键的。

题解：

直接forforfor瞎判断就好了……

只用判断每一个好的周围的那些路线就好了

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 120;
int mp[maxn][maxn][maxn];

int main()
{
    int n,m,k;
    scanf("%d%d%d",&n,&m,&k);
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            string s;cin>>s;
            for(int t=0;t<s.size();t++)
            {
                if(s[t]=='1')mp[i][j][t+1]=1;
                else mp[i][j][t+1]=0;
            }
        }
    }

    int ans = 0;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            for(int t=1;t<=k;t++)
            {
                if(mp[i][j][t])
                {
                    if(mp[i-1][j][t]&&mp[i+1][j][t])ans++;
                    else if(mp[i][j-1][t]&&mp[i][j+1][t])ans++;
                    else if(mp[i][j][t-1]&&mp[i][j][t+1])ans++;
                    else if(mp[i-1][j][t]&&mp[i][j+1][t]&&(!mp[i-1][j+1][t]))ans++;
                    else if(mp[i-1][j][t]&&mp[i][j][t+1]&&(!mp[i-1][j][t+1]))ans++;
                    else if(mp[i][j-1][t]&&mp[i+1][j][t]&&(!mp[i+1][j-1][t]))ans++;
                    else if(mp[i][j-1][t]&&mp[i][j][t+1]&&(!mp[i][j-1][t+1]))ans++;
                    else if(mp[i][j][t-1]&&mp[i+1][j][t]&&(!mp[i+1][j][t-1]))ans++;
                    else if(mp[i][j][t-1]&&mp[i][j+1][t]&&(!mp[i][j+1][t-1]))ans++;
                }
            }
        }
    }
    cout<<ans<<endl;
}