[HDU3062]Party（2-sat）

传送门

2-sat问题，只需要判断yes或no

所以可以直接连边，缩点，判断同一组的是否在同一个块中。

 #include <cstdio>
 #include <stack>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #define N 1000001

 int n, m, cnt, idx, sz;
 int head[N], to[N << ], next[N << ], dfn[N], low[N], belong[N];
 bool ins[N], flag;
 std::stack <int> s;

 inline void add(int x, int y)
 {
     to[cnt] = y;
     next[cnt] = head[x];
     head[x] = cnt++;
 }

 inline void tarjan(int u)
 {
     int i, v;
     dfn[u] = low[u] = ++idx;
     ins[u] = ;
     s.push(u);
     for(i = head[u]; i != -; i = next[i])
     {
         v = to[i];
         if(!dfn[v])
         {
             tarjan(v);
             low[u] = std::min(low[u], low[v]);
         }
         else if(ins[v]) low[u] = std::min(low[u], dfn[v]);
     }
     if(low[u] == dfn[u])
     {
         ++sz;
         do
         {
             v = s.top();
             s.pop();
             ins[v] = ;
             belong[v] = sz;
         }while(u != v);
     }
 }

 int main()
 {
     int i, j, a, b, c, d;
     while(~scanf("%d", &n))
     {
         memset(head, -, sizeof(head));
         memset(ins, , sizeof(ins));
         memset(dfn, , sizeof(dfn));
         idx = cnt = sz = ;
         while(!s.empty()) s.pop();
         scanf("%d", &m);
         for(i = ; i <= m; i++)
         {
             scanf("%d %d %d %d", &a, &b, &c, &d);
             a = (a << ) + c;
             b = (b << ) + d;
             add(a, b ^ );
             add(b, a ^ );
         }
         for(i = ; i < n << ; i++)
          if(!dfn[i])
           tarjan(i);
         flag = ;
         for(i = ; i < n; i++)
          if(belong[i << ] == belong[(i << ) ^ ])
           flag = ;
         if(flag) printf("NO\n");
         else printf("YES\n");
     }
     return ;
 }