hdu1814 Peaceful Commission

题目链接：戳我

菜得不行了，直到今天才刚开始学2-SAT。。。。

2-SAT的模板，因为是求最小字典序，所以只能用上限为\(O(nm)\)的最暴力的方法来做。。。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
#define MAXN 200010
using namespace std;
int n,m,t,cnt;
int head[MAXN],cur[MAXN],color[MAXN];
struct Edge{int nxt,to,dis;}edge[MAXN<<1];
inline int to(int x)
{if(x&1) return x+1;else return x-1;
}
inline void add(int from,int to){edge[++t].nxt=head[from],edge[t].to=to,head[from]=t;}
inline bool check(int x)
{if(color[x]&&color[x]==2) return false;else if(color[x]&&color[x]==1) return true;color[x]=1;color[to(x)]=2;cur[++cnt]=x;for(int i=head[x];i;i=edge[i].nxt)if(check(edge[i].to)==false) return false;return true;
}
inline bool solve()
{for(int i=1;i<=2*n;i++){if(color[i]) continue;cnt=0;if(check(i)==false){for(int j=1;j<=cnt;j++)color[cur[j]]=color[to(cur[j])]=0;if(check(to(i))==false) return false;}}return true;
}
int main()
{#ifndef ONLINE_JUDGEfreopen("ce.in","r",stdin);#endifwhile(scanf("%d%d",&n,&m)!=EOF){memset(edge,0,sizeof(edge));memset(head,0,sizeof(head));memset(color,0,sizeof(color));t=0;for(int i=1;i<=m;i++){int u,v;scanf("%d%d",&u,&v);add(u,to(v)),add(v,to(u));}if(solve()==true){for(int i=1;i<=2*n;i++)if(color[i]==1)printf("%d\n",i);}else printf("NIE\n");}return 0;
}