状态压缩 洛谷P2622 非DP

初学状态压缩 对位运算不太了解
传送门

//MADE BY Y_is_sunshine;
//#include <bits/stdc++.h>
//#include <memory.h>
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <cstdio>
#include <vector>
#include <string>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>

#define INF 0x3f3f3f3f
#define MAXN 400005

const int mod = 1e9 + 7;
const double PI = acos(-1);

using namespace std;

int f[105][15];
int step[MAXN];

int N, M;

void solve() {
    queue<int> q1;
    char s[15];
    for (int i = 0; i < N; i++)
        s[i] = '1';
    s[N] = '\0';
    q1.push(strtol(s, NULL, 2));
    step[strtol(s, NULL, 2)] = 1;
    while (!q1.empty()) {
        int k = q1.front();
        q1.pop();
        for (int i = 1; i <= M; i++) {
            int temp = k;
            for (int j = 1; j <= N; j++) {
                if (f[i][j] == 1) {
                    if(temp >> (j - 1) & 1)
                    temp = temp ^ (1 << (j - 1));
                }
                else if (f[i][j] == -1) {
                    temp = temp | (1 << (j - 1));
                }
            }
            if (!step[temp]) {

                step[temp] = step[k] + 1, q1.push(temp);
                if (!temp)
                    return;
            }

        }

    }
}

int main(void)
{
    freopen("data.txt", "r", stdin);

    ios::sync_with_stdio(false);
    //cin.tie(nullptr);
    cin.tie(NULL);


    cin >> N >> M;

    for (int i = 1; i <= M; i++) {
        for (int j = 1; j <= N; j++)
            cin >> f[i][j];
    }
    solve();

    if (step[0])
        cout << step[0] - 1 << endl;
    else
        cout << -1 << endl;


    freopen("CON", "r", stdin);
    system("pause");
    return 0;
}