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;

 

 1 #include <cstdio>
 2 #include <cstring>
 3 
 4 const int MAXN = 100010;
 5 
 6 int pa[MAXN];
 7 long long int d[MAXN];
 8 
 9 int findset( int x )
10 {
11     if ( pa[x] == x ) return x;
12     int root = findset( pa[x] );
13     d[x] += d[ pa[x] ];
14     return pa[x] = root;
15 }
16 
17 int main()
18 {
19     int N, M;
20     while ( scanf( "%d%d", &N, &M ), N || M )
21     {
22         for ( int i = 0; i <= N; ++i )
23         {
24             pa[i] = i;
25             d[i] = 0;
26         }
27         while ( M-- )
28         {
29             char op[4];
30             scanf( "%s", op );
31             int a, b, w;
32             if ( op[0] == '!' )
33             {
34                 scanf( "%d%d%d", &a, &b, &w );
35                 int u = findset(a);
36                 int v = findset(b);
37                 pa[v] = u;
38                 d[v] = d[a] - d[b] + w;
39             }
40             else
41             {
42                 scanf( "%d%d", &a, &b );
43                 int u = findset(a);
44                 int v = findset(b);
45                 if ( u != v ) puts("UNKNOWN");
46                 else printf( "%lld\n", d[b] - d[a] );
47             }
48         }
49     }
50     return 0;
51 }

 

posted @ 2013-07-15 17:31  冰鸮  阅读(300)  评论(0)  编辑  收藏  举报