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。

 

 1 #include <cstdio>

 2 #include <cstring>

 3 #include <cstdlib>

 4 

 5 const int MAXN = 200010;

 6 

 7 int N, Q;

 8 int p[MAXN];

 9 int d[MAXN];

10 

11 int FindSet( int x )

12 {

13     if ( p[x] == x ) return x;

14     int root = FindSet( p[x] );

15     d[x] += d[ p[x] ];

16     return p[x] = root;

17 }

18 

19 int main()

20 {

21     while ( ~scanf( "%d%d", &N, &Q ) )

22     {

23         for ( int i = 0; i <= N; ++i )

24         {

25             d[i] = 0;

26             p[i] = i;

27         }

28 

29         int cnt = 0;

30         while ( Q-- )

31         {

32             int a, b, sum;

33             scanf( "%d%d%d", &a, &b, &sum );

34             --a;

35             int u = FindSet( a );

36             int v = FindSet( b );

37             if ( u == v )

38             {

39                 if ( sum != d[b] - d[a] )

40                     ++cnt;

41             }

42             else

43             {

44                 p[v] = u;

45                 d[v] = d[a] - d[b] + sum;

46             }

47         }

48 

49         printf( "%d\n", cnt );

50     }

51     return 0;

52 }