【模板】 2-SAT 问题

2-SAT就是给出$m$个限制表示$x==val_x || y==val_y$

求出满足的解

每个点拆成两个点，如果$x$不满足则$y$一定满足，$y$不满足同理。这样我们连边，然后$tarjan$即可

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cstdlib>
 5 #include<cmath>
 6 #include<algorithm>
 7 #include<queue>
 8 #include<vector>
 9 #include<map>
10 #include<set>
11 #define ll long long
12 #define db double
13 #define inf 2139062143
14 #define MAXN 2001000
15 #define rep(i,s,t) for(register int i=(s),i##__end=(t);i<=i##__end;++i)
16 #define dwn(i,s,t) for(register int i=(s),i##__end=(t);i>=i##__end;--i)
17 #define ren for(register int i=fst[x];i;i=nxt[i])
18 #define pb(i,x) vec[i].push_back(x)
19 #define pls(a,b) (a+b)%MOD
20 #define mns(a,b) (a-b+MOD)%MOD
21 #define mul(a,b) (1LL*(a)*(b))%MOD
22 using namespace std;
23 inline int read()
24 {
25     int x=0,f=1;char ch=getchar();
26     while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();}
27     while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();}
28     return x*f;
29 }
30 int n,m,nxt[MAXN<<1],fst[MAXN],to[MAXN<<1],cnt;
31 int dfn[MAXN],low[MAXN],bl[MAXN],stp,st[MAXN],top,scc;
32 void add(int u,int v) {nxt[++cnt]=fst[u],fst[u]=cnt,to[cnt]=v;}
33 void tarjan(int x)
34 {
35     st[++top]=x,dfn[x]=low[x]=++stp;
36     ren if(!dfn[to[i]]) {tarjan(to[i]);low[x]=min(low[x],low[to[i]]);}
37         else if(!bl[to[i]]) low[x]=min(low[x],dfn[to[i]]);
38     if(low[x]==dfn[x])
39         {scc++;int now=0;while(now!=x) now=st[top--],bl[now]=scc;}
40 }
41 int main()
42 {
43     n=read(),m=read();int a,b,x,y;rep(i,1,m)
44         a=read(),x=read(),b=read(),y=read(),
45         add(a+(1^x)*n,b+y*n),add(b+(1^y)*n,a+x*n);
46     rep(i,1,n<<1) if(!dfn[i]) tarjan(i);
47     rep(i,1,n) if(bl[i]==bl[i+n]) return puts("IMPOSSIBLE"),0;puts("POSSIBLE");
48     rep(i,1,n) printf("%d ",bl[i]>bl[i+n]);
49 }

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