【BZOJ3531】【SDOI2014】旅行

题目传送门

题目大意：给定一棵无根树，每个节点有自己的类别和权值，现在给定两个类别相同的点，叫你求这2点路径上同类别节点的权值和／最大权值。

节点类别与权值会改变。

解题思路：考虑对每一个类别开一棵线段树，动态开点，可以写指针，也可以开数组写链表，然后剩下的就是树剖的东西了。我写的是指针，跑的略慢一些，时间效率为\( O(q \log^{2} n) \).

代码由于是在自家的mac上码的，所以和以前风格不大一样。

 #include <stdio.h>
 #define MN 100005
 inline void swp(int &a,int &b){a ^= b ^= a ^= b;}
 inline int max(int a,int b){return a > b ? a : b;}
 inline int in(){
     int x =  ,f = ; char ch = getchar();
     while (ch < '' || ch > '') f = ch=='-' ? - : , ch = getchar();
     while (ch >= '' && ch <= '' ) x = (x<<) + (x<<) + ch - '',ch = getchar();
     return x*f;
 }
 struct segment_tree{
     segment_tree *ls , *rs;
     int sum,ma;
     void combine(){
         sum = ma = ;
         if (ls != NULL){
             sum += ls -> sum;
             ma = max( ma , ls -> ma );
         }if (rs != NULL){
             sum += rs -> sum;
             ma = max( ma , rs -> ma );
         }
     }
     segment_tree (int sum=,int ma=):sum(sum),ma(ma){ls = rs = NULL;}
 }*ST[MN];
 #define S_T segment_tree
 int to[MN<<],nxt[MN<<],cnt,head[MN];
 int n,q,top[MN],siz[MN],son[MN],val[MN],belief[MN],fa[MN],dep[MN],pos[MN],dfsn;
 inline void ins(int x,int y){to[++cnt] = y , nxt[cnt] = head[x] , head[x] = cnt;}
 inline void insw(int x,int y){ins(x,y); ins(y,x);}
 inline void dfs1(int u,int f,int d){
     siz[u] = ; dep[u] = d; fa[u] = f;
     for (register int i = head[u]; i; i = nxt[i])
         if (to[i] != f) {
             dfs1(to[i],u,d+); siz[u] += siz[to[i]];
             if (siz[son[u]] < siz[to[i]]) son[u] = to[i];
         }
 }
 inline void dfs2(int u,int tp){
     top[u] = tp; pos[u] = (++dfsn); if (son[u]) dfs2(son[u],tp);
     for (register int i = head[u]; i; i = nxt[i])
         if (to[i] != fa[u] && to[i] != son[u]) dfs2(to[i],to[i]);
 }
 #define mid (l+r>>1)
 inline void insert(S_T* &x,int l,int r,int pos,int val){
     if (x == NULL) x = new S_T;
     if (l == r){x -> sum = x -> ma = val; return;}
     if (pos <= mid) insert(x -> ls,l,mid,pos,val);
     else insert(x -> rs,mid+,r,pos,val);x->combine();
 }
 inline void Delete(S_T* &x,int l,int r,int pos){
     if (l==r){delete x; x = NULL; return; }
     if (pos<=mid) Delete(x -> ls,l,mid,pos);
     else Delete(x -> rs,mid+,r,pos);
     if (x -> ls == NULL && x -> rs == NULL){delete x; x = NULL; return; }
     else x -> combine();
 }
 inline int QM(S_T* &x,int l,int r,int a,int b){
     if (x == NULL) return ;
     if (l == a && r == b) return x -> ma;
     if (b<=mid) return QM(x -> ls,l,mid,a,b);
     if (a>mid) return QM(x -> rs,mid+,r,a,b);
     return max(QM(x -> ls,l,mid,a,mid),QM(x -> rs,mid+,r,mid+,b));
 }
 inline int QS(S_T* &x,int l,int r,int a,int b){
     if (x == NULL) return ;
     if (l == a && r == b) return x -> sum;
     if (b<=mid) return QS(x -> ls,l,mid,a,b);
     if (a>mid) return QS(x -> rs,mid+,r,a,b);
     return QS(x -> ls,l,mid,a,mid)+QS(x -> rs,mid+,r,mid+,b);
 }
 inline int qm(int belief,int x,int y){
     register int res = ;
     while (top[x] != top[y]){
         if (dep[top[x]] < dep[top[y]]) swp(x,y);
         res = max(res,QM(ST[belief],,n,pos[top[x]],pos[x]));
         x = fa[top[x]];
     }if (dep[x] > dep[y]) swp(x,y);res = max(res,QM(ST[belief],,n,pos[x],pos[y]));
     return res;
 }
 inline int qs(int belief,int x,int y){
     register int res = ;
     while (top[x] != top[y]){
         if (dep[top[x]] < dep[top[y]]) swp(x,y);
         res += QS(ST[belief],,n,pos[top[x]],pos[x]);
         x = fa[top[x]];
     }if (dep[x] > dep[y]) swp(x,y);res += QS(ST[belief],,n,pos[x],pos[y]);
     return res;
 }
 void init(){
     n = in() , q = in();
     for (int i = ; i <= n; ++i) val[i] = in() , belief[i] = in();
     for (register int i = ; i < n; ++i) insw(in(),in());
     dfs1(,,);dfs2(,);
     for (register int i = ; i <= n; ++i)
         insert(ST[belief[i]],,n,pos[i],val[i]);
 }
 void solve(){
     while(q--){
         register char op[]; scanf("%s",op);
         register int x=in(),y=in();
         if (op[] == 'C'){
             if (op[] == 'C'){
                 Delete(ST[belief[x]],,n,pos[x]);
                 insert(ST[belief[x] = y],,n,pos[x],val[x]);
             }else insert(ST[belief[x]],,n,pos[x],val[x]=y);
         }else{
             if (op[] == 'M') printf("%d\n",qm(belief[x],x,y));
             else printf("%d\n",qs(belief[x],x,y));
         }
     }
 }
 int main(){ init(); solve();}