poj3207

题解：

2-sat

这次不是按照点来见图，而是按照遍

xi表示i号边在元内部，xi+m表示i在圆外部

然后判断一下两边在圆内圆外是否相交

代码：

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
using namespace std;
const int N=2005;
int flag[N],x[N],y[N],n,ne[N*N],fi[N],zz[N*N],num;
int t,zhan[N],dfn[N],m,l,ans,low[N],an[N];
void jb(int x,int y)
{
    ne[++num]=fi[x];
    fi[x]=num;
    zz[num]=y;
}
void dfs(int x)
{
    low[x]=dfn[x]=++l;
    zhan[++t]=x;
    flag[x]=true;
    for (int i=fi[x];i!=0;i=ne[i])
     {
         if (an[zz[i]])continue;
        if(!dfn[zz[i]])dfs(zz[i]);
        if(!flag[zz[i]])low[x]=min(low[x],dfn[zz[i]]);else
        low[x]=min(low[x],low[zz[i]]);
     }
    if (dfn[x]==low[x])
     {
         ans++;
         while (zhan[t]!=x)
          {
              flag[zhan[t]]=false;
              an[zhan[t--]]=ans;
          }
         an[zhan[t--]]=ans;
         flag[x]=false;
     }
}
int main()
{
    scanf("%d%d",&n,&m);
    for (int i=0;i<m;i++)
     {
         scanf("%d%d",&x[i],&y[i]);
         if (x[i]>y[i])swap(x[i],y[i]);
     }
    for (int i=0;i<m;i++)
     for (int j=i+1;j<m;j++)
      if (x[j]>=x[i]&&x[j]<=y[i]&&y[j]>=y[i]  
      ||y[j]>=x[i]&&y[j]<=y[i]&&x[j]<=x[i])
       {
           jb(i,j+m);
           jb(i+m,j);
           jb(j+m,i);
           jb(j,i+m);
       } 
    for (int i=0;i<2*m;i++)
     if (!dfn[i])dfs(i);
    for (int i=0;i<m;i++)
     if (an[i]==an[i+m])
      {
          puts("the evil panda is lying again");
          return 0;
      }
    puts("panda is telling the truth...");      
}