hdu 3062 2-SAT问题

思路：裸的2-SAT。

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<cstdio>
#include<vector>
#include<string>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define Maxn 2010
#define Maxm 100010
#define LL __int64
#define Abs(x) ((x)>0?(x):(-x))
#define lson(x) (x<<1)
#define rson(x) (x<<1|1)
#define inf 0x7fffffff
#define Mod 1000000007
using namespace std;
int dfn[Maxn],low[Maxn],id[Maxn],Stack[Maxn],lab,num,e,top,vi[Maxn],n,m;
vector<int> head[Maxn];
void init()
{
    memset(dfn,0,sizeof(dfn));
    memset(low,0,sizeof(low));
    lab=num=top=0;
    memset(vi,0,sizeof(vi));
    memset(id,0,sizeof(id));
    for(int i=0;i<=2*n;i++)
        head[i].clear();
}
void Tarjan(int u)
{
    int i,v,sz;
    dfn[u]=low[u]=++lab;
    Stack[top++]=u;
    vi[u]=1;
    sz=head[u].size();
    for(i=0;i<sz;i++){
        v=head[u][i];
        if(!dfn[v]){
            Tarjan(v);
            low[u]=min(low[v],low[u]);
        }
        if(vi[v])
        low[u]=min(low[u],dfn[v]);
    }
    if(low[u]==dfn[u]){
        ++num;
        do{
          i=Stack[--top];
          id[i]=num;
          vi[i]=0;
        }while(i!=u);
    }
    return ;
}
int main()
{
    int i,j,a,b,c,d;
    while(scanf("%d",&n)!=EOF){
        init();
        scanf("%d",&m);
        for(i=1;i<=m;i++){
            scanf("%d%d%d%d",&a,&b,&c,&d);
            a++;
            b++;
            if(c==0&&d==0)
                head[a].push_back(b+n),head[b].push_back(a+n);
            if(c==0&&d==1)
                head[a].push_back(b),head[b+n].push_back(a+n);
            if(c==1&&d==0)
                head[a+n].push_back(b+n),head[b].push_back(a);
            if(c==1&&d==1)
                head[a+n].push_back(b),head[b+n].push_back(a);
        }
        for(i=1;i<=2*n;i++){
            if(!dfn[i])
                Tarjan(i);
        }
        for(i=1;i<=n;i++)
            if(id[i]==id[i+n])
                break;
        if(i<=n)
            printf("NO
");
        else
            printf("YES
");
    }
    return 0;
}