poj3678

题解：

2-sat问题

构图十分巧妙

具体可以看我的代码的构图部分

代码：

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
using namespace std;
const int N=2005;
char s[5];
int flag[N],x,y,z,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);
    while (m--)
     {
         scanf("%d%d%d%s",&x,&y,&z,&s);
         if (s[0]=='A'&&z==0)jb(x+n,y),jb(y+n,x);
         if (s[0]=='A'&&z==1)jb(x,x+n),jb(y,y+n);
         if (s[0]=='O'&&z==0)jb(x+n,x),jb(y+n,y);
         if (s[0]=='O'&&z==1)jb(x,y+n),jb(y,x+n);
         if (s[0]=='X'&&z==0)jb(x+n,y+n),jb(x,y),jb(y+n,x+n),jb(y,x);
         if (s[0]=='X'&&z==1)jb(x+n,y),jb(x,y+n),jb(y+n,x),jb(y,x+n);
     }
    for (int i=0;i<2*n;i++)
     if (!dfn[i])dfs(i);
    for (int i=0;i<n;i++)
     if (an[i]==an[i+n])
      {
          puts("NO");
          return 0;
      }  
    puts("YES");
    return 0;  
}