hdu1814

题解：

2-sat

nm暴力

虽然似乎复杂度最低

代码：

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
using namespace std;
const int N=;
int num,ne[N],fi[N],n,m,x,y,f[N],zz[N],vis[N],S[N],top;
void jb(int x,int y)
{
    ne[++num]=fi[x];
    fi[x]=num;
    zz[num]=y;
}
int dfs(int x)
{
    if (vis[x^])return ;
    if (vis[x])return ;
    vis[x]=;
    S[top++]=x;
    for (int i=fi[x];i;i=ne[i])
     if (!dfs(zz[i]))return ;
    return ;
}
int slove()
{
    memset(vis,,sizeof vis);
    for (int i=;i<*n;i+=)
     {
         if (vis[i]||vis[i^])continue;
          top=;
          if (!dfs(i))
           {
               while (top)vis[S[--top]]=;
               if (!dfs(i^))return ;
           }
     }
    return ;
}
int main()
{
    while (~scanf("%d%d",&n,&m))
      {
          memset(fi,,sizeof fi);
          num=;
        while (m--)
          {
             scanf("%d%d",&x,&y);
             x--;y--;
             jb(x,y^);jb(y,x^);
          }
        if (slove())
          {
             for (int i=;i<*n;i++)
               if (vis[i])printf("%d\n",i+);
          }
        else puts("NIE");
     }
    return ;
}