Luo Gu -P2262- lights problem II (bfs + state compression Preliminary)

Title Description

N conventional lamps, and m buttons. Each button can simultaneously control the lamp n - i-button is pressed, all lamps have an effect. For the first pressed i j lamp, is one of the 3 following effects: If A [i] [j] is 1, then when this lamp when opened, shut it off, otherwise no matter; If -1 then, if this lamp is off, then open it, otherwise no matter; if it is 0, regardless of whether these lights on, do not care.

Now these lights are open, given all the switches to control the effect of all the lamps, inquire at least a few Yaoan button to switch off all.

Input Format

The first two rows of two numbers, nm

Subsequently m rows, each row number n, a [i] [j] indicates the effect of the i-th switch of j-th light.

Output Format

An integer representing the minimum number press the button. If there is no way to make it all off, output -1

Input # 1

3
2
1 0 1
-1 1 0

Output # 1

2

Problem-solving ideas:

This problem can be searched first thought, bfs, enumeration affect the n switches for each state.
Then for this question in a state, it is known, each point only two states, i.e., on or off, then we can use the number of binary bits to represent a light switch, that is to say for the i th light, if the i the number of positions is then i say this lamp is open, otherwise closed.
A state of the modifier may also be used if the nature of the bit operation, bit i of a number of modifications.
So write good code, then dfs indicates the status binary search, the search is on to explain the state found a solution 0:00.
In fact, that is the state of compression, it is no compression ah, nothing more than a status value represents a binary number. That white is simple violence.

AC Code:

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N = 1e6 +10;
struct node {
    int sta,step;
    node(int a,int b)
    {
        sta = a;
        step = b;
    }
};

int a[105][105];
int n,m;
bool mark[1<<11];

int bfs()
{
    int sta = 1;
    for(int i = 1 ; i < m ; i ++)
    {
        sta |= 1<<i;
    }
    memset(mark,0,sizeof(mark));
    queue<node> que;
    que.push(node(sta,0));
    while(!que.empty())
    {
        node tmp = que.front();
        que.pop();
        if(tmp.sta == 0)
            return tmp.step;
        for(int i = 0 ; i < n ; i ++)
        {
            int tmp1 = tmp.sta;
            for(int j = 0 ; j < m ; j ++)
            {
                if(a[i][j] == 1 && (tmp.sta&(1<<j)))
                    tmp1 = tmp1^(1<<j);
                if(a[i][j] == -1)
                    tmp1 = tmp1|(1<<j);
            }
            if(!mark[tmp1])
                que.push(node(tmp1,tmp.step+1));
            mark[tmp1] = true;
        }
    }
    return -1;
}

signed main()
{
    cin>>m>>n;
    for(int i = 0 ; i < n ; i ++)
        for(int j = 0 ; j < m ; j ++)
            cin>>a[i][j];
    cout<<bfs()<<endl;
}