Luogu3676 小清新数据结构题 动态点分治

Posted 2021-01-25 itst

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换根类型的统计问题动态点分治都是很好做的。

设所有点的点权和为$sum$

首先，我们先不考虑求$sumlimits_i s_i^2$，先考虑如何在换根的情况下求$sumlimits_i s_i$。

考虑一个点$i$会被统计多少次，显然是$dep_i+1$，那么$sumlimits_i s_i = sumlimits_i (dep_i+1) imes val_i = sumlimits_i dep_i imes val_i + sum$。

$sumlimits_i dep_i imes val_i$是不是似曾相识……其实就是幻想乡战略游戏

接着我们考虑这个烦人的平方项。那么我们还需要推一个结论：换根不会影响$W=sumlimits_i s_i imes (sum - s_i)$的值。

证明如下：

我们考虑$K = sumlimits_i sumlimits_j val_i imes val_j imes dis(i,j)$，意思就是取两个点，将中间的所有边设为两边的点的权值之积然后相加。显然这一个值是不会因为根的变化而改变的。

接着我们考虑一条边$(x,y)$对$K$的贡献，显然是这条边分割开来的两个子树的权值和的乘积，而无论根是哪一个，这两个子树中必定有且仅有一个是树的一个子树，假设$x$对应的树是子树，它就代表着$s_x imes (sum - s_x)$，把所有边的贡献加起来就是$W=sumlimits_i s_i imes (sum - s_i)=sumlimits_i sumlimits_j val_i imes val_j imes dis(i,j)$，所以$W$是不会因为根的变化而变化的。

而$W=sumlimits_i s_i imes (sum - s_i) = sum imes sumlimits_i s_i - sumlimits_i s_i^2$，那么$sumlimits_i s_i^2 = sum imes sumlimits_i s_i - W$

接着我们考虑修改点权对$W$的影响。由$Delta W = Delta v imes sumlimits_j val_j imes dep_j$，实质跟上面计算$sumlimits_i s_i$的方法一样。

  1 #include<bits/stdc++.h>
  2 #define int long long
  3 //This code is written by Itst
  4 using namespace std;
  5 
  6 inline int read(){
  7     int a = 0;
  8     bool f = 0;
  9     char c = getchar();
 10     while(c != EOF && !isdigit(c)){
 11         if(c == ‘-‘)
 12             f = 1;
 13         c = getchar();
 14     }
 15     while(c != EOF && isdigit(c)){
 16         a = (a << 3) + (a << 1) + (c ^ ‘0‘);
 17         c = getchar();
 18     }
 19     return f ? -a : a;
 20 }
 21 
 22 const int MAXN = 200010;
 23 struct Edge{
 24     int end , upEd;
 25 }Ed[MAXN << 1];
 26 int val[MAXN] , head[MAXN] , fa[MAXN][20] , dis[MAXN][20] , dep[MAXN];
 27 int ST[21][MAXN << 1] , fir[MAXN] , logg2[MAXN << 1] , size[MAXN] , cur[MAXN] , up[MAXN] , sum[MAXN];
 28 int cntEd , N , M , minSize , nowSize , minInd , cntST , W , allV;
 29 bool vis[MAXN];
 30 
 31 inline void addEd(int a , int b){
 32     Ed[++cntEd].end = b;
 33     Ed[cntEd].upEd = head[a];
 34     head[a] = cntEd;
 35 }
 36 
 37 void init_dfs(int x , int p){
 38     dep[x] = dep[p] + 1;
 39     fir[x] = ++cntST;
 40     sum[x] = val[x];
 41     ST[0][cntST] = x;
 42     for(int i = head[x] ; i ; i = Ed[i].upEd)
 43         if(Ed[i].end != p){
 44             init_dfs(Ed[i].end , x);
 45             ST[0][++cntST] = x;
 46             sum[x] += sum[Ed[i].end];
 47         }
 48     W += sum[x] * (allV - sum[x]);
 49 }
 50 
 51 inline int cmp(int a , int b){
 52     return dep[a] < dep[b] ? a : b;
 53 }
 54 
 55 void init_st(){
 56     for(int i = 2 ; i <= cntST ; ++i)
 57         logg2[i] = logg2[i >> 1] + 1;
 58     for(int i = 1 ; 1 << i <= cntST ; ++i)
 59         for(int j = 1 ; j + (1 << i) - 1 <= cntST ; ++j)
 60             ST[i][j] = cmp(ST[i - 1][j] , ST[i - 1][j + (1 << (i - 1))]);
 61 }
 62 
 63 inline int LCA(int x , int y){
 64     x = fir[x];
 65     y = fir[y];
 66     if(x > y)
 67         swap(x , y);
 68     int t = logg2[y - x + 1];
 69     return cmp(ST[t][x] , ST[t][y - (1 << t) + 1]);
 70 }
 71 
 72 inline int calcLen(int x , int y){
 73     return dep[x] + dep[y] - (dep[LCA(x , y)] << 1);
 74 }
 75 
 76 void getSize(int x){
 77     vis[x] = 1;
 78     ++nowSize;
 79     for(int i = head[x] ; i ; i = Ed[i].upEd)
 80         if(!vis[Ed[i].end])
 81             getSize(Ed[i].end);
 82     vis[x] = 0;
 83 }
 84 
 85 void getRoot(int x){
 86     vis[x] = size[x] = 1;
 87     int maxN = 0;
 88     for(int i = head[x] ; i ; i = Ed[i].upEd)
 89         if(!vis[Ed[i].end]){
 90             getRoot(Ed[i].end);
 91             maxN = max(maxN , size[Ed[i].end]);
 92             size[x] += size[Ed[i].end];
 93         }
 94     maxN = max(maxN , nowSize - size[x]);
 95     if(maxN < minSize){
 96         minSize = maxN;
 97         minInd = x;
 98     }
 99     vis[x] = 0;
100 }
101 
102 void init_dfz(int x , int p){
103     nowSize = 0;
104     minSize = 0x7fffffff;
105     getSize(x);
106     getRoot(x);
107     x = minInd;
108     fa[x][0] = x;
109     fa[x][1] = p;
110     vis[x] = 1;
111     sum[x] = val[x];
112     for(int i = 1 ; fa[x][i] ; ++i){
113         fa[x][i + 1] = fa[fa[x][i]][1];
114         dis[x][i] = calcLen(fa[x][i] , x);
115         up[fa[x][i - 1]] += dis[x][i] * val[x];
116         sum[fa[x][i]] += val[x];
117         cur[fa[x][i]] += dis[x][i] * val[x];
118     }
119     for(int i = head[x] ; i ; i = Ed[i].upEd)
120         if(!vis[Ed[i].end])
121             init_dfz(Ed[i].end , x);
122 }
123 
124 void init(){
125     init_dfs(1 , 0);
126     init_st();
127     init_dfz(1 , 0);
128 }
129 
130 inline int query(int x){
131     int all = cur[x];
132     for(int i = 1 ; fa[x][i] ; ++i){
133         all += cur[fa[x][i]] - up[fa[x][i - 1]];
134         all += (sum[fa[x][i]] - sum[fa[x][i - 1]]) * dis[x][i];
135     }
136     return all;
137 }
138 
139 inline void modify(int x , int v){
140     W += (v - val[x]) * query(x);
141     allV += v - val[x];
142     sum[x] += v - val[x];
143     for(int i = 1 ; fa[x][i] ; ++i){
144         up[fa[x][i - 1]] += (v - val[x]) * dis[x][i];
145         cur[fa[x][i]] += (v - val[x]) * dis[x][i];
146         sum[fa[x][i]] += v - val[x];
147     }
148     val[x] = v;
149 }
150 
151 signed main(){
152 #ifndef ONLINE_JUDGE
153     freopen("3676.in" , "r" , stdin);
154     freopen("3676.out" , "w" , stdout);
155 #endif
156     N = read();
157     M = read();
158     for(int i = 1 ; i < N ; ++i){
159         int a = read() , b = read();
160         addEd(a , b);
161         addEd(b , a);
162     }
163     for(int i = 1 ; i <= N ; ++i){
164         val[i] = read();
165         allV += val[i];
166     }
167     init();
168     for(int i = 1 ; i <= M ; ++i)
169         if(read() == 1){
170             int x = read() , y = read();
171             modify(x , y);
172         }
173         else
174             cout << allV * (query(read()) + allV) - W << ‘
‘;
175     return 0;
176 }