LA 6187 - Never Wait for Weights 并查集的带权路径压缩

只有一个地方需要注意：

设节点a的根为u，b的跟为v，则：a = u + d[a];  b = v + d[b];

已知：b-a=w。所以v - u = d[a] - d[b] + w;

在合并两个集合修改根节点时，把v的根改为u，同时v到根的距离为d[a] - d[b] + w;

 #include <cstdio>
 #include <cstring>

 const int MAXN = ;

 int pa[MAXN];
 long long int d[MAXN];

 int findset( int x )
 {
     if ( pa[x] == x ) return x;
     int root = findset( pa[x] );
     d[x] += d[ pa[x] ];
     return pa[x] = root;
 }

 int main()
 {
     int N, M;
     while ( scanf( "%d%d", &N, &M ), N || M )
     {
         for ( int i = ; i <= N; ++i )
         {
             pa[i] = i;
             d[i] = ;
         }
         while ( M-- )
         {
             char op[];
             scanf( "%s", op );
             int a, b, w;
             if ( op[] == '!' )
             {
                 scanf( "%d%d%d", &a, &b, &w );
                 int u = findset(a);
                 int v = findset(b);
                 pa[v] = u;
                 d[v] = d[a] - d[b] + w;
             }
             else
             {
                 scanf( "%d%d", &a, &b );
                 int u = findset(a);
                 int v = findset(b);
                 if ( u != v ) puts("UNKNOWN");
                 else printf( "%lld\n", d[b] - d[a] );
             }
         }
     }
     return ;
 }