【模板】洛谷P2622_状态压缩+bfs_关灯问题_

该题数据较小，想到状态压缩
不可能则输出-1，想到bfs队空后的输出状态
每个状态都是由上个状态到达的

#include <cstdio>
#include <queue>
#include <cstring>

using namespace std;

int n, m, a[111][11];
bool vis[1<<11];
typedef struct pii {int x, s;}pii;//x:状态, s:当前步数
queue <pii> q;

int bfs()
{
    pii u, v;
    int i, j, t;
    while(!q.empty())
    {
        u = q.front();
        q.pop();
        v.s = u.s + 1;
        for(i = 0; i < m; ++i)//枚举灯
        {
            t = u.x;
            for(j = 0; j < n; ++j)//用灯i分别操作t状态的n盏灯
                if((a[i][j]==1 && t&(1<<j)) || (a[i][j]==-1 && !(t&(1<<j))))
                    t ^= (1<<j);
            if(!t) return v.s;
            if(vis[t]) continue;
            v.x = t;
            vis[t] = 1;
            q.push(v);
        }
    }
    return -1;
}

int main()
{
    scanf("%d %d", &n, &m);
    for(int i = 0; i < m; ++i)
        for(int j = 0; j < n; ++j)
            scanf("%d", &a[i][j]);
    pii p;
    p.x = (1<<n)-1;
    p.s = 0;
    q.push(p);
    memset(vis, 0, sizeof(vis));
    vis[p.x] = 1;
    int ans = bfs();
    printf("%d\n", ans);
    return 0;
}