hdu1814

Posted 2020-10-18 宣毅鸣

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=200005;
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^1])return 0;
    if (vis[x])return 1;
    vis[x]=1;
    S[top++]=x;
    for (int i=fi[x];i;i=ne[i])
     if (!dfs(zz[i]))return 0;
    return 1; 
}
int slove()
{
    memset(vis,0,sizeof vis);
    for (int i=0;i<2*n;i+=2)
     {
         if (vis[i]||vis[i^1])continue;
          top=0;
          if (!dfs(i))
           {
               while (top)vis[S[--top]]=0;
               if (!dfs(i^1))return 0;
           }
     }
    return 1;  
}
int main()
{
    while (~scanf("%d%d",&n,&m))
      {
          memset(fi,0,sizeof fi);
          num=0;
        while (m--)
          {
             scanf("%d%d",&x,&y);
             x--;y--;
             jb(x,y^1);jb(y,x^1);
          }
        if (slove())
          {
             for (int i=0;i<2*n;i++)
               if (vis[i])printf("%d\n",i+1);
          }
        else puts("NIE"); 
     }
    return 0; 
}