The gist of the title: Given the equality or inequality relationship of n variables with each other, find out whether these relationships are contradictory
Idea: Add equal variables to the set and check whether the unequal query is legal
eg: The data is large and discretized (however I use map)
#include<stdio.h> #include<algorithm> #include<string.h> #include<map> using namespace std; int fa[10000000],n,sum[10000000],tot,T; map<int,int> old,tong; struct node { int x,y,c; }a[10000000]; template<class T>void read(T &x) { x=0;char ch=getchar(); while(ch<'0'||ch>'9') ch=getchar(); while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();} } int find(int x) { int p=x,q; while(x!=fa[x]) x=fa[x]; while(p!=x) { q = fa [p]; fa [p] = x; p = q } return x; } intmain () { read(T); while(T--) { old.clear();tong.clear();tot=0;bool flag=0; read(n); for(int i=1;i<=n;i++) { read(a[i].x);read(a[i].y);read(a[i].c); if(tong[a[i].x]==0) { tong[a[i].x]=1; sum [ ++ tot] = a [i] .x; } if(tong[a[i].y]==0) { tong[a[i].y]=1; sum [ ++ tot] = a [i] .y; } } sort(sum+1,sum+1+tot); for(int i=1;i<=tot;i++) old[sum[i]]=i; for(int i=1;i<=n;i++) { a[i].x=old[a[i].x]; a[i].y=old[a[i].y]; } for(int i=1;i<=tot;i++) fa[i]=i; for(int i=1;i<=n;i++) { int p=find(a[i].x),q=find(a[i].y); if(a[i].c==1&&p!=q) fa[p]=q; if(a[i].c!=1&&p==q){flag=1;break;} } if(flag==1){printf("NO\n");continue;} for(int i=1;i<=n;i++) { int p=find(a[i].x),q=find(a[i].y); if(a[i].c==1&&p!=q){flag=1;break;} if(a[i].c==0&&p==q){flag=1;break;} } if(flag==1)printf("NO\n"); else printf("YES\n"); } return 0; }