BZOJ5210 最大连通子块和 【树链剖分】【堆】【动态DP】

题目分析：

解决了上次提到的《切树游戏》后，这道题就是一道模板题。

注意我们需要用堆维护子重链的最大值。这样不会使得复杂度变坏，因为每个重链我们只考虑一个点。

时间复杂度$O(nlog^2n)$

代码：

 #include<bits/stdc++.h>
 using namespace std;
 
 typedef long long ll;
 
 const int maxn = ;
 
 int n,m;
 int v[maxn];
 vector<int> g[maxn];
 int fa[maxn],dep[maxn],sz[maxn],son[maxn],top[maxn],tail[maxn];
 int where[maxn],number[maxn],num;
 ll TOT[maxn];
 
 struct Priority_queue{
     priority_queue <ll,vector<ll>,less<ll> > pq,del;
     void Insert(ll now){pq.push(now);}
     void Erase(ll now){del.push(now);}
     ll Top(){
     while(!del.empty()&&pq.top() == del.top()) pq.pop(),del.pop();
     return pq.top();
     }
     int Size(){return pq.size()-del.size();}
 }PQ[maxn];
 
 struct node{ll L,R,D,C;}T[maxn<<];
 
 void push_up(int now){
     T[now].L = max(T[now<<].L,T[now<<].C+T[now<<|].L);
     T[now].R = max(T[now<<|].R,T[now<<|].C+T[now<<].R);
     T[now].D = max(max(T[now<<].D,T[now<<|].D),T[now<<].R+T[now<<|].L);
     T[now].C = T[now<<].C+T[now<<|].C;
 }
 
 node merge(node alpha,node beta){
     node gamma; gamma.L = max(alpha.L,alpha.C+beta.L);
     gamma.R = max(beta.R,beta.C+alpha.R);
     gamma.D = max(max(alpha.D,beta.D),alpha.R+beta.L);
     gamma.C = alpha.C+beta.C;
     return gamma;
 }
 
 char readchar(){
     char ch = getchar(); while(ch != 'M' && ch != 'Q') ch = getchar();
     return ch;
 }
 
 void read(){
     scanf("%d%d",&n,&m);
     for(int i=;i<=n;i++) scanf("%d",&v[i]);
     for(int i=;i<n;i++){
     int x,y; scanf("%d%d",&x,&y);
     g[x].push_back(y); g[y].push_back(x);
     }
 }
 
 void dfs1(int now,int f,int dp){
     fa[now] = f; dep[now] = dp;
     for(int i=;i<g[now].size();i++){
     if(g[now][i] == f) continue;
     dfs1(g[now][i],now,dp+);
     sz[now] += sz[g[now][i]];
     if(son[now]==||sz[son[now]]<sz[g[now][i]])son[now]=g[now][i];
     }
     sz[now]++;
     if(son[now]) tail[now] = tail[son[now]];
     else tail[now] = now;
 }
 
 void dfs2(int now,int tp){
     top[now] = tp;number[now] = ++num; where[num] = now;
     if(son[now]) dfs2(son[now],tp);
     for(int i=;i<g[now].size();i++){
     if(g[now][i] == fa[now] || g[now][i] == son[now]) continue;
     dfs2(g[now][i],g[now][i]);
     }
 }
 
 void Modify(int now,int tl,int tr,int place){
     if(tl == tr){
     tl = where[tl];
     T[now].C = TOT[tl] + v[tl];
     T[now].L = max(0ll,TOT[tl]+v[tl]); T[now].R = T[now].L;
     if(PQ[tl].Size()) T[now].D = max(PQ[tl].Top(),T[now].L);
     else T[now].D = T[now].L;
     }else{
     int mid = (tl+tr)/;
     if(place <= mid) Modify(now<<,tl,mid,place);
     else Modify(now<<|,mid+,tr,place);
     push_up(now);
     }
 }
 
 node Query(int now,int tl,int tr,int l,int r){
     if(tl >= l && tr <= r) return T[now];
     int mid = (tl+tr)/;
     if(r <= mid) return Query(now<<,tl,mid,l,r);
     if(l > mid) return Query(now<<|,mid+,tr,l,r);
     return merge(Query(now<<,tl,mid,l,r),Query(now<<|,mid+,tr,l,r));
 }
 
 void dfs3(int now){
     if(son[now]) dfs3(son[now]);
     for(int i=;i<g[now].size();i++){
     int mp = g[now][i];
     if(mp == fa[now] || mp == son[now]) continue;
     dfs3(mp);
     PQ[now].Insert(Query(,,n,number[mp],number[tail[mp]]).D);
     }
     Modify(,,n,number[now]);
     if(top[now] == now){
     long long data = Query(,,n,number[now],number[tail[now]]).L;
     if(data >  && fa[now]) TOT[fa[now]]+=data;
     }
 }
 
 void work(){
     dfs1(,,);
     dfs2(,);
     dfs3();
     for(int i=;i<=m;i++){
     char ch = readchar();
     if(ch == 'M'){
         int x,y; scanf("%d%d",&x,&y);
         stack<int> sta; int now = top[x];
         while(fa[now]) sta.push(now),now = top[fa[now]];
         while(!sta.empty()){
         int mp = sta.top();sta.pop();
         node res = Query(,,n,number[mp],number[tail[mp]]);
         PQ[fa[mp]].Erase(res.D);
         if(res.L > ) TOT[fa[mp]]-=res.L;
         Modify(,,n,number[fa[mp]]);
         }
         now = x; v[x] = y;
         while(now){
         Modify(,,n,number[now]);
         now = top[now];
         node res = Query(,,n,number[now],number[tail[now]]);
         if(res.L >  && fa[now]) TOT[fa[now]]+=res.L;
         PQ[fa[now]].Insert(res.D);
         now= fa[now];
         }
     }else{
         int x; scanf("%d",&x);
         long long ans = Query(,,n,number[x],number[tail[x]]).D;
         printf("%lld\n",ans);
     }
     }
 }
 
 int main(){
     read();
     work();
     return ;
 }