poj1733 Parity game[带权并查集or扩展域]

地址

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连通性判定问题。（具体参考lyd并查集专题该题的转化方法，反正我菜我没想出来）。转化后就是一个经典的并查集问题了。

带权：要求两点奇偶性不同，即连边权为1，否则为0，压缩路径时不断异或，可以通过0或1得到两点的关系。合并时解一个位运算的方程，可得一个根连向另一个根的权值，看code，就不细讲了。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<algorithm>
 #include<queue>
 #define dbg(x) cerr<<#x<<" = "<<x<<endl
 #define ddbg(x,y) cerr<<#x<<" = "<<x<<"   "<<#y<<" = "<<y<<endl
 using namespace std;
 typedef long long ll;
 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 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 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=+;
 struct kishin_sagume{
     int l,r,p;
 }q[N];
 int a[N<<],fa[N<<],d[N<<];
 int L,n,m,x,y,fx,fy,flag,i;
 inline int Get(int x){
     if(fa[x]==x)return x;
     int ret=Get(fa[x]);d[x]^=d[fa[x]];
     return fa[x]=ret;
 }

 int main(){//freopen("test.in","r",stdin);//freopen("test.out","w",stdout);
     read(L),read(n);char s[];
     for(i=;i<=n;++i){
         read(q[i].l),read(q[i].r),a[++m]=--q[i].l,a[++m]=q[i].r;
         scanf("%s",s);q[i].p=s[]=='o';
     }
     sort(a+,a+m+),m=unique(a+,a+m+)-a-;
     for(i=;i<=m;++i)fa[i]=i;
     for(i=;i<=n;++i){
         x=lower_bound(a+,a+m+,q[i].l)-a,y=lower_bound(a+,a+m+,q[i].r)-a;
         fx=Get(x),fy=Get(y);
         if(fx^fy)fa[fx]=fy,d[fx]=q[i].p^d[x]^d[y];
         else if(d[x]^d[y]^q[i].p)break;
     }
     return printf("%d\n",i-),;
 }

扩展域：同奇偶时连奇数域和偶数域分别连边，否则交错，边的含义是可以产生关系，或者说可以推出。已知他是满足传递性的。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<algorithm>
 #include<queue>
 #define dbg(x) cerr<<#x<<" = "<<x<<endl
 #define ddbg(x,y) cerr<<#x<<" = "<<x<<"   "<<#y<<" = "<<y<<endl
 using namespace std;
 typedef long long ll;
 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 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 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=+;
 struct matara_okina{
     int l,r,p;
 }q[N];
 int a[N<<],fa[N<<];
 int L,n,m,x,y,i;
 inline int Get(int x){
     return fa[x]^x?fa[x]=Get(fa[x]):x;
 }

 int main(){//freopen("test.in","r",stdin);//freopen("test.out","w",stdout);
     read(L),read(n);char s[];
     for(i=;i<=n;++i){
         read(q[i].l),read(q[i].r),a[++m]=--q[i].l,a[++m]=q[i].r;
         scanf("%s",s);q[i].p=s[]=='o';
     }
     sort(a+,a+m+),m=unique(a+,a+m+)-a-;
     for(i=;i<=(m<<);++i)fa[i]=i;
     for(i=;i<=n;++i){
         x=lower_bound(a+,a+m+,q[i].l)-a,y=lower_bound(a+,a+m+,q[i].r)-a;
         if(q[i].p){
             if(Get(x)==Get(y))break;
             fa[Get(x)]=Get(y+m),fa[Get(x+m)]=Get(y);
         }
         else{
             if(Get(x)==Get(y+m))break;
             fa[Get(x)]=Get(y),fa[Get(x+m)]=Get(y+m);
         }
     }
     return printf("%d\n",i-),;
 }