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：

1 #include<bits/stdc++.h>
2 using namespace std;
3
4 typedef long long ll;
5
6 const int maxn = 202000;
7
8 int n,m;
9 int v[maxn];
10 vector<int> g[maxn];
11 int fa[maxn],dep[maxn],sz[maxn],son[maxn],top[maxn],tail[maxn];
12 int where[maxn],number[maxn],num;
13 ll TOT[maxn];
14
15 struct Priority_queue{
16     priority_queue <ll,vector<ll>,less<ll> > pq,del;
17     void Insert(ll now){pq.push(now);}
18     void Erase(ll now){del.push(now);}
19     ll Top(){
20     while(!del.empty()&&pq.top() == del.top()) pq.pop(),del.pop();
21     return pq.top();
22     }
23     int Size(){return pq.size()-del.size();}
24 }PQ[maxn];
25
26 struct node{ll L,R,D,C;}T[maxn<<2];
27
28 void push_up(int now){
29     T[now].L = max(T[now<<1].L,T[now<<1].C+T[now<<1|1].L);
30     T[now].R = max(T[now<<1|1].R,T[now<<1|1].C+T[now<<1].R);
31     T[now].D = max(max(T[now<<1].D,T[now<<1|1].D),T[now<<1].R+T[now<<1|1].L);
32     T[now].C = T[now<<1].C+T[now<<1|1].C;
33 }
34
35 node merge(node alpha,node beta){
36     node gamma; gamma.L = max(alpha.L,alpha.C+beta.L);
37     gamma.R = max(beta.R,beta.C+alpha.R);
38     gamma.D = max(max(alpha.D,beta.D),alpha.R+beta.L);
39     gamma.C = alpha.C+beta.C;
40     return gamma;
41 }
42
43 char readchar(){
44     char ch = getchar(); while(ch != 'M' && ch != 'Q') ch = getchar();
45     return ch;
46 }
47
48 void read(){
49     scanf("%d%d",&n,&m);
50     for(int i=1;i<=n;i++) scanf("%d",&v[i]);
51     for(int i=1;i<n;i++){
52     int x,y; scanf("%d%d",&x,&y);
53     g[x].push_back(y); g[y].push_back(x);
54     }
55 }
56
57 void dfs1(int now,int f,int dp){
58     fa[now] = f; dep[now] = dp;
59     for(int i=0;i<g[now].size();i++){
60     if(g[now][i] == f) continue;
61     dfs1(g[now][i],now,dp+1);
62     sz[now] += sz[g[now][i]];
63     if(son[now]==0||sz[son[now]]<sz[g[now][i]])son[now]=g[now][i];
64     }
65     sz[now]++;
66     if(son[now]) tail[now] = tail[son[now]];
67     else tail[now] = now;
68 }
69
70 void dfs2(int now,int tp){
71     top[now] = tp;number[now] = ++num; where[num] = now;
72     if(son[now]) dfs2(son[now],tp);
73     for(int i=0;i<g[now].size();i++){
74     if(g[now][i] == fa[now] || g[now][i] == son[now]) continue;
75     dfs2(g[now][i],g[now][i]);
76     }
77 }
78
79 void Modify(int now,int tl,int tr,int place){
80     if(tl == tr){
81     tl = where[tl];
82     T[now].C = TOT[tl] + v[tl];
83     T[now].L = max(0ll,TOT[tl]+v[tl]); T[now].R = T[now].L;
84     if(PQ[tl].Size()) T[now].D = max(PQ[tl].Top(),T[now].L);
85     else T[now].D = T[now].L;
86     }else{
87     int mid = (tl+tr)/2;
88     if(place <= mid) Modify(now<<1,tl,mid,place);
89     else Modify(now<<1|1,mid+1,tr,place);
90     push_up(now);
91     }
92 }
93
94 node Query(int now,int tl,int tr,int l,int r){
95     if(tl >= l && tr <= r) return T[now];
96     int mid = (tl+tr)/2;
97     if(r <= mid) return Query(now<<1,tl,mid,l,r);
98     if(l > mid) return Query(now<<1|1,mid+1,tr,l,r);
99     return merge(Query(now<<1,tl,mid,l,r),Query(now<<1|1,mid+1,tr,l,r));
100 }
101
102 void dfs3(int now){
103     if(son[now]) dfs3(son[now]);
104     for(int i=0;i<g[now].size();i++){
105     int mp = g[now][i];
106     if(mp == fa[now] || mp == son[now]) continue;
107     dfs3(mp);
108     PQ[now].Insert(Query(1,1,n,number[mp],number[tail[mp]]).D);
109     }
110     Modify(1,1,n,number[now]);
111     if(top[now] == now){
112     long long data = Query(1,1,n,number[now],number[tail[now]]).L;
113     if(data > 0 && fa[now]) TOT[fa[now]]+=data;
114     }
115 }
116
117 void work(){
118     dfs1(1,0,1);
119     dfs2(1,1);
120     dfs3(1);
121     for(int i=1;i<=m;i++){
122     char ch = readchar();
123     if(ch == 'M'){
124         int x,y; scanf("%d%d",&x,&y);
125         stack<int> sta; int now = top[x];
126         while(fa[now]) sta.push(now),now = top[fa[now]];
127         while(!sta.empty()){
128         int mp = sta.top();sta.pop();
129         node res = Query(1,1,n,number[mp],number[tail[mp]]);
130         PQ[fa[mp]].Erase(res.D);
131         if(res.L > 0) TOT[fa[mp]]-=res.L;
132         Modify(1,1,n,number[fa[mp]]);
133         }
134         now = x; v[x] = y;
135         while(now){
136         Modify(1,1,n,number[now]);
137         now = top[now];
138         node res = Query(1,1,n,number[now],number[tail[now]]);
139         if(res.L > 0 && fa[now]) TOT[fa[now]]+=res.L;
140         PQ[fa[now]].Insert(res.D);
141         now= fa[now];
142         }
143     }else{
144         int x; scanf("%d",&x);
145         long long ans = Query(1,1,n,number[x],number[tail[x]]).D;
146         printf("%lld\n",ans);
147     }
148     }
149 }
150
151 int main(){
152     read();
153     work();
154     return 0;
155 }