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 = 200010;

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 = 0; i <= N; ++i )
        {
            d[i] = 0;
            p[i] = i;
        }

        int cnt = 0;
        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
", cnt );
    }
    return 0;
}