这题转化一下题意就是给一堆形如$a_i + a_j e cquad (a_iin [0,1],cin [0,2])$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。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<cmath> 6 #include<queue> 7 #define mst(x) memset(x,0,sizeof x) 8 #define dbg(x) cerr << #x << " = " << x <<endl 9 #define dbg2(x,y) cerr<< #x <<" = "<< x <<" "<< #y <<" = "<< y <<endl 10 using namespace std; 11 typedef long long ll; 12 typedef double db; 13 typedef pair<int,int> pii; 14 template<typename T>inline T _min(T A,T B){return A<B?A:B;} 15 template<typename T>inline T _max(T A,T B){return A>B?A:B;} 16 template<typename T>inline char MIN(T&A,T B){return A>B?(A=B,1):0;} 17 template<typename T>inline char MAX(T&A,T B){return A<B?(A=B,1):0;} 18 template<typename T>inline void _swap(T&A,T&B){A^=B^=A^=B;} 19 template<typename T>inline T read(T&x){ 20 x=0;int f=0;char c;while(!isdigit(c=getchar()))if(c=='-')f=1; 21 while(isdigit(c))x=x*10+(c&15),c=getchar();return f?x=-x:x; 22 } 23 const int N=400+7,M=10000+7; 24 struct stothx{int x,y,c;}Q[M]; 25 int n,m,T; 26 struct thxorz{ 27 int head[N],to[M<<2],nxt[M<<2],dfn[N],low[N],tim,stk[N],instk[N],bel[N],Top,scc,tot; 28 inline void clear(){mst(head),mst(dfn),tot=tim=scc=0;} 29 inline void link(int x,int y){to[++tot]=y,nxt[tot]=head[x],head[x]=tot;} 30 void tarjan(int x){//dbg(x); 31 #define y to[j] 32 dfn[x]=low[x]=++tim,stk[++Top]=x,instk[x]=1; 33 for(register int j=head[x];j;j=nxt[j]){ 34 if(!dfn[y])tarjan(y),MIN(low[x],low[y]); 35 else if(instk[y])MIN(low[x],dfn[y]); 36 } 37 if(dfn[x]==low[x]){ 38 int tmp;++scc; 39 do instk[tmp=stk[Top--]]=0,bel[tmp]=scc;while(tmp^x); 40 } 41 #undef y 42 } 43 inline bool check(){ 44 for(register int i=1;i<=n;++i)if(bel[i]==bel[i+n])return 0; 45 return 1; 46 } 47 }G; 48 inline bool check(int mid){//dbg(mid); 49 G.clear(); 50 for(register int i=1;i<=mid;++i){ 51 int x=Q[i].x,y=Q[i].y,c=Q[i].c; 52 if(x==y){ 53 if(c==0)G.link(x,x+n); 54 else if(c==2)G.link(x+n,x); 55 } 56 else{ 57 if(c==0)G.link(x,y+n),G.link(y,x+n); 58 else if(c==1)G.link(x,y),G.link(y,x),G.link(x+n,y+n),G.link(y+n,x+n); 59 else G.link(x+n,y),G.link(y+n,x); 60 } 61 } 62 for(register int i=1;i<=n<<1;++i)if(!G.dfn[i])G.tarjan(i); 63 return G.check(); 64 } 65 66 int main(){//freopen("test.in","r",stdin);//freopen("test.ans","w",stdout); 67 read(T);while(T--){ 68 read(n),read(m); 69 for(register int i=1;i<=m;++i)read(Q[i].x),read(Q[i].y),read(Q[i].c),++Q[i].x,++Q[i].y; 70 int L=0,R=m; 71 while(L<R){ 72 int mid=L+R+1>>1; 73 if(check(mid))L=mid; 74 else R=mid-1; 75 } 76 printf("%d ",L); 77 } 78 return 0; 79 }
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。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<cmath> 6 #include<queue> 7 #define mst(x) memset(x,0,sizeof x) 8 #define dbg(x) cerr << #x << " = " << x <<endl 9 #define dbg2(x,y) cerr<< #x <<" = "<< x <<" "<< #y <<" = "<< y <<endl 10 using namespace std; 11 typedef long long ll; 12 typedef double db; 13 typedef pair<int,int> pii; 14 template<typename T>inline T _min(T A,T B){return A<B?A:B;} 15 template<typename T>inline T _max(T A,T B){return A>B?A:B;} 16 template<typename T>inline char MIN(T&A,T B){return A>B?(A=B,1):0;} 17 template<typename T>inline char MAX(T&A,T B){return A<B?(A=B,1):0;} 18 template<typename T>inline void _swap(T&A,T&B){A^=B^=A^=B;} 19 template<typename T>inline T read(T&x){ 20 x=0;int f=0;char c;while(!isdigit(c=getchar()))if(c=='-')f=1; 21 while(isdigit(c))x=x*10+(c&15),c=getchar();return f?x=-x:x; 22 } 23 const int N=3000+7,M=3000+7; 24 struct stothx{int x,y,c;}Q[M]; 25 int n,m,T; 26 struct thxorz{ 27 int head[N],to[M<<1],nxt[M<<1],dfn[N],low[N],tim,stk[N],instk[N],bel[N],Top,scc,tot; 28 inline void clear(){mst(head),mst(dfn),tot=tim=scc=0;} 29 inline void link(int x,int y){to[++tot]=y,nxt[tot]=head[x],head[x]=tot;} 30 void tarjan(int x){//dbg(x); 31 #define y to[j] 32 dfn[x]=low[x]=++tim,stk[++Top]=x,instk[x]=1; 33 for(register int j=head[x];j;j=nxt[j]){ 34 if(!dfn[y])tarjan(y),MIN(low[x],low[y]); 35 else if(instk[y])MIN(low[x],dfn[y]); 36 } 37 if(dfn[x]==low[x]){ 38 int tmp;++scc; 39 do instk[tmp=stk[Top--]]=0,bel[tmp]=scc;while(tmp^x); 40 } 41 #undef y 42 } 43 inline bool check(){ 44 for(register int i=1;i<=n;++i)if(bel[i]==bel[i+n])return 0; 45 return 1; 46 } 47 }G; 48 int bel[N],id[N],a[M],b[M]; 49 inline bool check(int mid){//dbg(mid); 50 G.clear(); 51 for(register int i=1;i<=mid;++i){ 52 int x=bel[a[i]],y=bel[b[i]],xi=id[a[i]],yi=id[b[i]]; 53 if(a[i]==b[i])xi?G.link(x,x+n):G.link(x+n,x); 54 else if(x^y)G.link(xi?x:x+n,yi?y+n:y),G.link(yi?y:y+n,xi?x+n:x); 55 } 56 for(register int i=1;i<=n<<1;++i)if(!G.dfn[i])G.tarjan(i); 57 return G.check(); 58 } 59 60 int main(){//freopen("test.in","r",stdin);//freopen("test.ans","w",stdout); 61 while(read(n),read(m),n||m){ 62 for(register int i=1,x,y;i<=n;++i){ 63 read(x),read(y); 64 bel[x]=bel[y]=i,id[x]=0,id[y]=1; 65 } 66 for(register int i=1;i<=m;++i)read(a[i]),read(b[i]); 67 int L=0,R=m; 68 while(L<R){ 69 int mid=L+R+1>>1; 70 if(check(mid))L=mid; 71 else R=mid-1; 72 } 73 printf("%d ",L); 74 } 75 return 0; 76 }