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
", cnt );
50     }
51     return 0;
52 }