2020蓝桥杯七段码简单好理解BFS解法

#include <bits/stdc++.h>

using namespace std;
int m[10][10];
//从起点开始进行bfs，有边可以走，就走，
//然后把该边置为0，ans++，把每次到达的点丢入队列
//看最后能走几条边

int bfs(int st)
{
    int ans = 0;
    queue<int> Q;
    Q.push(st);
    while (!Q.empty())
    {
        int now = Q.front();
        Q.pop();
        for (int i = 1; i <= 7; i++)
        {
            if (m[now][i])
            {
                m[now][i] = m[i][now] = 0;
                Q.push(i);
                ans++;
            }
        }
    }
    return ans;
}

int main()
{
    int res = 0;
    for (int i = 1; i < 1 << 8; i++)
    {
        memset(m, 0, sizeof(m));
        int x = i;
        int loc = 1;
        int st=0, cnt = 0;
        while (x)
        {
            int v = x % 2;
            if (v)
            {
                cnt++;
                if (loc < 7)
                    st = loc;
                else
                    st = 3;
            }
            switch (loc)
            {
            case 1:
                m[1][2] = m[2][1] = v;
                break;
            case 2:
                m[2][4] = m[4][2] = v;
                break;
            case 3:
                m[3][1] = m[1][3] = v;
                break;
            case 4:
                m[4][3] = m[3][4] = v;
                break;
            case 5:
                m[5][6] = m[6][5] = v;
                break;
            case 6:
                m[6][4] = m[4][6] = v;
                break;
            case 7:
                m[3][5] = m[5][3] = v;
                break;
            default:
                break;
            }
            x /= 2;
            loc++;
        }
        if (bfs(st) == cnt)
            res++;
    }
    cout << res << endl;
    return 0;
}
//答案80