bzoj 3306

以1号节点为根，弄出DFS序，我们发现，对于一个询问：(rt,u)，以rt为根，u节点的子树中的最小点权，我们可以根据rt,u,1这三个节点在同一条路径上的相对关系来把它转化为以1为根的在DFS序上的区间询问（中间有一种情况要在树上倍增，理解了LCA的话应该很容易写出来）。

收获：

对于只有换根这种改变树的形态的操作，又只询问和子树相关的问题，可以不用动态树。

  1 /**************************************************************
  2     Problem: 3306
  3     User: idy002
  4     Language: C++
  5     Result: Accepted
  6     Time:2264 ms
  7     Memory:19948 kb
  8 ****************************************************************/
  9
 10 #include <cstdio>
 11 #include <iostream>
 12 #define oo 0x3f3f3f3f
 13 #define N 100010
 14 #define P 16
 15 using namespace std;
 16
 17 struct Node {
 18     int v;
 19     Node *ls, *rs;
 20 }pool[N*3], *tail=pool, *root;
 21
 22 int n, m;
 23 int head[N], wght[N], dest[N+N], next[N+N], etot;
 24 int anc[N][P+1], dep[N], in[N], out[N], vdf[N], idc;
 25
 26 void update( Node *nd ) {
 27     nd->v = min( nd->ls->v, nd->rs->v );
 28 }
 29 Node *build( int lf, int rg ) {
 30     Node *nd = ++tail;
 31     if( lf==rg ) {
 32         nd->v = wght[vdf[lf]];
 33         return nd;
 34     }
 35     int mid=(lf+rg)>>1;
 36     nd->ls = build( lf, mid );
 37     nd->rs = build( mid+1, rg );
 38     update( nd );
 39     return nd;
 40 }
 41 void modify( Node *nd, int lf, int rg, int pos, int val ) {
 42     if( lf==rg ) {
 43         nd->v = val;
 44         return;
 45     }
 46     int mid=(lf+rg)>>1;
 47     if( pos<=mid ) modify( nd->ls, lf, mid, pos, val );
 48     else modify( nd->rs, mid+1, rg, pos, val );
 49     update(nd);
 50 }
 51 int query( Node *nd, int lf, int rg, int L, int R ) {
 52     if( L<=lf && rg<=R )
 53         return nd->v;
 54     int mid=(lf+rg)>>1;
 55     int rt = oo;
 56     if( L<=mid ) rt = query( nd->ls, lf, mid, L, R );
 57     if( R>mid ) rt = min( rt, query( nd->rs, mid+1, rg, L, R ) );
 58     return rt;
 59 }
 60 void adde( int u, int v ) {
 61     etot++;
 62     next[etot] = head[u];
 63     dest[etot] = v;
 64     head[u] = etot;
 65 }
 66 void dfs( int u ) {
 67     ++idc;
 68     in[u] = idc;
 69     vdf[idc] = u;
 70     for( int p=1; p<=P; p++ )
 71         anc[u][p] = anc[anc[u][p-1]][p-1];
 72     for( int t=head[u]; t; t=next[t] ) {
 73         int v=dest[t];
 74         anc[v][0] = u;
 75         dep[v] = dep[u]+1;
 76         dfs(v);
 77     }
 78     out[u] = idc;
 79 }
 80 int climb( int u, int t ) {
 81     for( int p=0; t; t>>=1,p++ )
 82         if( t&1 ) u=anc[u][p];
 83     return u;
 84 }
 85 int lca( int u, int v ) {
 86     if( dep[u]<dep[v] ) swap(u,v);
 87     int t=dep[u]-dep[v];
 88     u = climb( u, t );
 89     if( u==v ) return u;
 90     for( int p=P; p>=0&&anc[u][0]!=anc[v][0]; p-- )
 91         if( anc[u][p]!=anc[v][p] ) u=anc[u][p],v=anc[v][p];
 92     return anc[u][0];
 93 }
 94 int query( int rt, int u ) {
 95     int ca=lca(rt,u);
 96     if( rt==u ) {
 97         return query( root, 1, idc, 1, idc );
 98     } else if( ca==u ) {
 99         int ans1=oo, ans2=oo;
100         u = climb( rt, dep[rt]-dep[u]-1 );
101         if( in[u]>=2 ) ans1 = query( root, 1, idc, 1, in[u]-1 );
102         if( out[u]<=n-1 ) ans2 = query( root, 1, idc, out[u]+1, idc );
103         return min( ans1, ans2 );
104     } else {
105         return query( root, 1, idc, in[u], out[u] );
106     }
107 }
108 int main() {
109     scanf( "%d%d", &n, &m );
110     for( int i=1,f,w; i<=n; i++ ) {
111         scanf( "%d%d", &f, &w );
112         wght[i] = w;
113         if( f ) adde(f,i);
114     }
115     anc[1][0] = 1;
116     dep[1] = 1;
117     dfs(1);
118     root = build( 1, idc );
119
120     int crt=1;
121     for( int t=1,x,y; t<=m; t++ ) {
122         char ch[10];
123         scanf( "%s", ch );
124
125         if( ch[0]==‘V‘ ) {
126             scanf( "%d%d", &x, &y );
127             modify( root, 1, idc, in[x], y );
128         } else if( ch[0]==‘E‘ ) {
129             scanf( "%d", &crt );
130         } else {
131             scanf( "%d", &x );
132             printf( "%d\n", query(crt,x) );
133         }
134     }
135 }