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 = 202000;

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<<2];

void push_up(int now){
    T[now].L = max(T[now<<1].L,T[now<<1].C+T[now<<1|1].L);
    T[now].R = max(T[now<<1|1].R,T[now<<1|1].C+T[now<<1].R);
    T[now].D = max(max(T[now<<1].D,T[now<<1|1].D),T[now<<1].R+T[now<<1|1].L);
    T[now].C = T[now<<1].C+T[now<<1|1].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=1;i<=n;i++) scanf("%d",&v[i]);
    for(int i=1;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=0;i<g[now].size();i++){
    if(g[now][i] == f) continue;
    dfs1(g[now][i],now,dp+1);
    sz[now] += sz[g[now][i]];
    if(son[now]==0||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=0;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)/2;
    if(place <= mid) Modify(now<<1,tl,mid,place);
    else Modify(now<<1|1,mid+1,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)/2;
    if(r <= mid) return Query(now<<1,tl,mid,l,r);
    if(l > mid) return Query(now<<1|1,mid+1,tr,l,r);
    return merge(Query(now<<1,tl,mid,l,r),Query(now<<1|1,mid+1,tr,l,r));
}

void dfs3(int now){
    if(son[now]) dfs3(son[now]);
    for(int i=0;i<g[now].size();i++){
    int mp = g[now][i];
    if(mp == fa[now] || mp == son[now]) continue;
    dfs3(mp);
    PQ[now].Insert(Query(1,1,n,number[mp],number[tail[mp]]).D);
    }
    Modify(1,1,n,number[now]);
    if(top[now] == now){
    long long data = Query(1,1,n,number[now],number[tail[now]]).L;
    if(data > 0 && fa[now]) TOT[fa[now]]+=data; 
    }
}

void work(){
    dfs1(1,0,1);
    dfs2(1,1);
    dfs3(1);
    for(int i=1;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(1,1,n,number[mp],number[tail[mp]]);
        PQ[fa[mp]].Erase(res.D);
        if(res.L > 0) TOT[fa[mp]]-=res.L;
        Modify(1,1,n,number[fa[mp]]);
        }
        now = x; v[x] = y;
        while(now){
        Modify(1,1,n,number[now]);
        now = top[now];
        node res = Query(1,1,n,number[now],number[tail[now]]);
        if(res.L > 0 && 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(1,1,n,number[x],number[tail[x]]).D;
        printf("%lld
",ans);
    }
    }
}

int main(){
    read();
    work();
    return 0;
}