HDU 3038 How Many Answers Are Wrong 并查集带权路径压缩

思路跟 LA 6187 完全一样。

我是乍一看没反应过来这是个并查集，知道之后就好做了。

d[i]代表节点 i 到根节点的距离，即每次的sum。

 #include <cstdio>
 #include <cstring>
 #include <cstdlib>

 const int MAXN = ;

 int N, Q;
 int p[MAXN];
 int d[MAXN];

 int FindSet( int x )
 {
     if ( p[x] == x ) return x;
     int root = FindSet( p[x] );
     d[x] += d[ p[x] ];
     return p[x] = root;
 }

 int main()
 {
     while ( ~scanf( "%d%d", &N, &Q ) )
     {
         for ( int i = ; i <= N; ++i )
         {
             d[i] = ;
             p[i] = i;
         }

         int cnt = ;
         while ( Q-- )
         {
             int a, b, sum;
             scanf( "%d%d%d", &a, &b, &sum );
             --a;
             int u = FindSet( a );
             int v = FindSet( b );
             if ( u == v )
             {
                 if ( sum != d[b] - d[a] )
                     ++cnt;
             }
             else
             {
                 p[v] = u;
                 d[v] = d[a] - d[b] + sum;
             }
         }

         printf( "%d\n", cnt );
     }
     return ;
 }