CodeForces 776D 2-SAT

CodeForces 776D    

题意：n个开关，m个房间，每个房间一定由两个开关控制，给出初始房间状态，问是否能把房间全部打开。

tags：因为每个房间一定由两个开关控制，把开关当变量，房间初始状态当条件，按2-SAT建边，然后dfs判断即可。

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define rep(i,a,b) for (int i=a;i<=b;i++)
#define per(i,b,a) for (int i=b;i>=a;i--)
#define mes(a,b)  memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define MP make_pair
#define PB push_back
#define fi  first
#define se  second
typedef long long ll;
const int N = 200005;

int n, m, room[N], a[N], b[N];
bool vis[N<<1];
vector<int > G[N<<1];
void Addedge(int u, int v) { G[u].PB(v); G[v].PB(u); }
int sta[N<<2], top;
bool dfs(int u)
{
    if(vis[u^1]) return false;
    if(vis[u]) return true;
    vis[u]=true, sta[top++]=u;
    for(auto v : G[u]) if(!dfs(v)) return false;
    return true;
}
int main()
{
    scanf("%d %d", &n, &m);
    rep(i,1,n) scanf("%d", &room[i]);
    int xi, ai;
    rep(i,1,m)
    {
        scanf("%d", &xi);
        rep(j,1,xi)
        {
            scanf("%d", &ai);
            if(a[ai]==0) a[ai]=i;
            else  b[ai]=i;
        }
    }
    rep(i,1,n)
    {
        int x=a[i], y=b[i];
        if(room[i]==0) Addedge(2*x, 2*y+1), Addedge(2*x+1, 2*y);
        else Addedge(2*x, 2*y), Addedge(2*x+1, 2*y+1);
    }
    mes(vis, false);
    for(int i=2; i<=m*2; i+=2)
    {
        if(vis[i]==false && vis[i+1]==false)
        {
            top=0;
            if(!dfs(i)) {
                while(top>0) vis[sta[--top]]=false;
                if(!dfs(i+1))  return 0*printf("NO");
            }
        }
    }
    puts("YES");

    return 0;
}