hdu3715 Go Deeper[二分+2-SAT]/poj2723 Get Luffy Out[二分+2-SAT]

这题转化一下题意就是给一堆形如$a_i + a_j \ne c\quad (a_i\in [0,1],c\in [0,2])$的限制，问从开头开始最多到哪条限制全是有解的。

那么，首先有可二分性，所以直接二分枚举最大处，然后把这些限制加边做一次2-sat就好了。连边的话注意一下细节就行，$c=0$时候就是选$0$必须另一个选$1$，$c=1$是选同类，$c=2$就是选$1$另一个必须选$0$，然后注意一下$i=j$也就是对同一变量的特殊讨论，就是直接赋值型的连边。然后就没了。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<queue>
 #define mst(x) memset(x,0,sizeof x)
 #define dbg(x) cerr << #x << " = " << x <<endl
 #define dbg2(x,y) cerr<< #x <<" = "<< x <<"  "<< #y <<" = "<< y <<endl
 using namespace std;
 typedef long long ll;
 typedef double db;
 typedef pair<int,int> pii;
 template<typename T>inline T _min(T A,T B){return A<B?A:B;}
 template<typename T>inline T _max(T A,T B){return A>B?A:B;}
 template<typename T>inline char MIN(T&A,T B){return A>B?(A=B,):;}
 template<typename T>inline char MAX(T&A,T B){return A<B?(A=B,):;}
 template<typename T>inline void _swap(T&A,T&B){A^=B^=A^=B;}
 template<typename T>inline T read(T&x){
     x=;int f=;char c;while(!isdigit(c=getchar()))if(c=='-')f=;
     while(isdigit(c))x=x*+(c&),c=getchar();return f?x=-x:x;
 }
 const int N=+,M=+;
 struct stothx{int x,y,c;}Q[M];
 int n,m,T;
 struct thxorz{
     int head[N],to[M<<],nxt[M<<],dfn[N],low[N],tim,stk[N],instk[N],bel[N],Top,scc,tot;
     inline void clear(){mst(head),mst(dfn),tot=tim=scc=;}
     inline void link(int x,int y){to[++tot]=y,nxt[tot]=head[x],head[x]=tot;}
     void tarjan(int x){//dbg(x);
         #define y to[j]
         dfn[x]=low[x]=++tim,stk[++Top]=x,instk[x]=;
         for(register int j=head[x];j;j=nxt[j]){
             if(!dfn[y])tarjan(y),MIN(low[x],low[y]);
             else if(instk[y])MIN(low[x],dfn[y]);
         }
         if(dfn[x]==low[x]){
             int tmp;++scc;
             do instk[tmp=stk[Top--]]=,bel[tmp]=scc;while(tmp^x);
         }
         #undef y
     }
     inline bool check(){
         for(register int i=;i<=n;++i)if(bel[i]==bel[i+n])return ;
         return ;
     }
 }G;
 inline bool check(int mid){//dbg(mid);
     G.clear();
     for(register int i=;i<=mid;++i){
         int x=Q[i].x,y=Q[i].y,c=Q[i].c;
         if(x==y){
             if(c==)G.link(x,x+n);
             else if(c==)G.link(x+n,x);
         }
         else{
             if(c==)G.link(x,y+n),G.link(y,x+n);
             else if(c==)G.link(x,y),G.link(y,x),G.link(x+n,y+n),G.link(y+n,x+n);
             else G.link(x+n,y),G.link(y+n,x);
         }
     }
     for(register int i=;i<=n<<;++i)if(!G.dfn[i])G.tarjan(i);
     return G.check();
 }

 int main(){//freopen("test.in","r",stdin);//freopen("test.ans","w",stdout);
     read(T);while(T--){
         read(n),read(m);
         for(register int i=;i<=m;++i)read(Q[i].x),read(Q[i].y),read(Q[i].c),++Q[i].x,++Q[i].y;
         int L=,R=m;
         while(L<R){
             int mid=L+R+>>;
             if(check(mid))L=mid;
             else R=mid-;
         }
         printf("%d\n",L);
     }
     return ;
 }

hdu3715

然后poj这题基本思路基本都是差不多，就是建边的时候花样变了一下，具体情况3类分析一下即可，同样注意特殊情况（同状态连边）要考虑到。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<queue>
 #define mst(x) memset(x,0,sizeof x)
 #define dbg(x) cerr << #x << " = " << x <<endl
 #define dbg2(x,y) cerr<< #x <<" = "<< x <<"  "<< #y <<" = "<< y <<endl
 using namespace std;
 typedef long long ll;
 typedef double db;
 typedef pair<int,int> pii;
 template<typename T>inline T _min(T A,T B){return A<B?A:B;}
 template<typename T>inline T _max(T A,T B){return A>B?A:B;}
 template<typename T>inline char MIN(T&A,T B){return A>B?(A=B,):;}
 template<typename T>inline char MAX(T&A,T B){return A<B?(A=B,):;}
 template<typename T>inline void _swap(T&A,T&B){A^=B^=A^=B;}
 template<typename T>inline T read(T&x){
     x=;int f=;char c;while(!isdigit(c=getchar()))if(c=='-')f=;
     while(isdigit(c))x=x*+(c&),c=getchar();return f?x=-x:x;
 }
 const int N=+,M=+;
 struct stothx{int x,y,c;}Q[M];
 int n,m,T;
 struct thxorz{
     int head[N],to[M<<],nxt[M<<],dfn[N],low[N],tim,stk[N],instk[N],bel[N],Top,scc,tot;
     inline void clear(){mst(head),mst(dfn),tot=tim=scc=;}
     inline void link(int x,int y){to[++tot]=y,nxt[tot]=head[x],head[x]=tot;}
     void tarjan(int x){//dbg(x);
         #define y to[j]
         dfn[x]=low[x]=++tim,stk[++Top]=x,instk[x]=;
         for(register int j=head[x];j;j=nxt[j]){
             if(!dfn[y])tarjan(y),MIN(low[x],low[y]);
             else if(instk[y])MIN(low[x],dfn[y]);
         }
         if(dfn[x]==low[x]){
             int tmp;++scc;
             do instk[tmp=stk[Top--]]=,bel[tmp]=scc;while(tmp^x);
         }
         #undef y
     }
     inline bool check(){
         for(register int i=;i<=n;++i)if(bel[i]==bel[i+n])return ;
         return ;
     }
 }G;
 int bel[N],id[N],a[M],b[M];
 inline bool check(int mid){//dbg(mid);
     G.clear();
     for(register int i=;i<=mid;++i){
         int x=bel[a[i]],y=bel[b[i]],xi=id[a[i]],yi=id[b[i]];
         if(a[i]==b[i])xi?G.link(x,x+n):G.link(x+n,x);
         else if(x^y)G.link(xi?x:x+n,yi?y+n:y),G.link(yi?y:y+n,xi?x+n:x);
     }
     for(register int i=;i<=n<<;++i)if(!G.dfn[i])G.tarjan(i);
     return G.check();
 }

 int main(){//freopen("test.in","r",stdin);//freopen("test.ans","w",stdout);
     while(read(n),read(m),n||m){
         for(register int i=,x,y;i<=n;++i){
             read(x),read(y);
             bel[x]=bel[y]=i,id[x]=,id[y]=;
         }
         for(register int i=;i<=m;++i)read(a[i]),read(b[i]);
         int L=,R=m;
         while(L<R){
             int mid=L+R+>>;
             if(check(mid))L=mid;
             else R=mid-;
         }
         printf("%d\n",L);
     }
     return ;
 }

poj2723