CodeForces 383C Propagating tree
Propagating tree
64-bit integer IO format: %I64d Java class name: (Any)
This tree has a special property: when a value val is added to a value of node i, the value -val is added to values of all the children of node i. Note that when you add value -val to a child of node i, you also add -(-val) to all children of the child of node i and so on. Look an example explanation to understand better how it works.
This tree supports two types of queries:
- "1 x val" — val is added to the value of node x;
- "2 x" — print the current value of node x.
In order to help Iahub understand the tree better, you must answer m queries of the preceding type.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 200000). The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000). Each of the next n–1 lines contains two integers viand ui (1 ≤ vi, ui ≤ n), meaning that there is an edge between nodes vi and ui.
Each of the next m lines contains a query in the format described above. It is guaranteed that the following constraints hold for all queries: 1 ≤ x ≤ n, 1 ≤ val ≤ 1000.
Output
For each query of type two (print the value of node x) you must print the answer to the query on a separate line. The queries must be answered in the order given in the input.
Sample Input
5 5
1 2 1 1 2
1 2
1 3
2 4
2 5
1 2 3
1 1 2
2 1
2 2
2 4
3
3
0
Hint
The values of the nodes are [1, 2, 1, 1, 2] at the beginning.
Then value 3 is added to node 2. It propagates and value -3 is added to it's sons, node 4 and node 5. Then it cannot propagate any more. So the values of the nodes are [1, 5, 1, - 2, - 1].
Then value 2 is added to node 1. It propagates and value -2 is added to it's sons, node 2 and node 3. From node 2 it propagates again, adding value 2 to it's sons, node 4 and node 5. Node 3 has no sons, so it cannot propagate from there. The values of the nodes are [3, 3, - 1, 0, 1].
You can see all the definitions about the tree at the following link: http://en.wikipedia.org/wiki/Tree_(graph_theory)
Source
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <climits> 7 #include <vector> 8 #include <queue> 9 #include <cstdlib> 10 #include <string> 11 #include <set> 12 #include <stack> 13 #define LL long long 14 #define pii pair<int,int> 15 #define INF 0x3f3f3f3f 16 using namespace std; 17 const int maxn = 300010; 18 struct node { 19 int lt,rt,val; 20 }; 21 struct arc{ 22 int to,next; 23 arc(int x = 0,int y = -1){ 24 to = x; 25 next = y; 26 } 27 }; 28 node tree[2][maxn<<2]; 29 arc e[maxn<<2]; 30 int head[maxn],lt[maxn],rt[maxn],val[maxn],d[maxn],idx,tot,n,m; 31 void add(int u,int v){ 32 e[tot] = arc(v,head[u]); 33 head[u] = tot++; 34 } 35 void dfs(int u,int fa){ 36 lt[u] = ++idx; 37 d[u] = d[fa]+1; 38 for(int i = head[u]; ~i; i = e[i].next){ 39 if(e[i].to == fa) continue; 40 dfs(e[i].to,u); 41 } 42 rt[u] = idx; 43 } 44 void build(int lt,int rt,int v) { 45 tree[0][v].lt = tree[1][v].lt = lt; 46 tree[0][v].rt = tree[1][v].rt = rt; 47 tree[0][v].val = tree[1][v].val = 0; 48 if(lt == rt) return; 49 int mid = (lt + rt)>>1; 50 build(lt,mid,v<<1); 51 build(mid+1,rt,v<<1|1); 52 } 53 void pushdown(int v,int o) { 54 if(tree[o][v].val) { 55 tree[o][v<<1].val += tree[o][v].val; 56 tree[o][v<<1|1].val += tree[o][v].val; 57 tree[o][v].val = 0; 58 } 59 } 60 void update(int lt,int rt,int v,int val,int o) { 61 if(tree[o][v].lt >= lt && tree[o][v].rt <= rt) { 62 tree[o][v].val += val; 63 return; 64 } 65 pushdown(v,o); 66 if(lt <= tree[o][v<<1].rt) update(lt,rt,v<<1,val,o); 67 if(rt >= tree[o][v<<1|1].lt) update(lt,rt,v<<1|1,val,o); 68 } 69 int query(int p,int v,int o){ 70 if(tree[o][v].lt == tree[o][v].rt) 71 return tree[o][v].val; 72 pushdown(v,o); 73 if(p <= tree[o][v<<1].rt) return query(p,v<<1,o); 74 if(p >= tree[o][v<<1|1].lt) return query(p,v<<1|1,o); 75 } 76 int main() { 77 int u,v,op; 78 while(~scanf("%d %d",&n,&m)){ 79 memset(head,-1,sizeof(head)); 80 idx = tot = 0; 81 for(int i = 1; i <= n; ++i) 82 scanf("%d",val+i); 83 for(int i = 1; i < n; ++i){ 84 scanf("%d %d",&u,&v); 85 add(u,v); 86 add(v,u); 87 } 88 build(1,n,1); 89 d[0] = 0; 90 dfs(1,0); 91 for(int i = 0; i < m; ++i){ 92 scanf("%d",&op); 93 if(op == 1){ 94 scanf("%d %d",&u,&v); 95 update(lt[u],rt[u],1,v,d[u]&1); 96 if(lt[u] < rt[u]) update(lt[u]+1,rt[u],1,-v,(~d[u])&1); 97 }else{ 98 scanf("%d",&u); 99 printf("%d\n",val[u]+query(lt[u],1,d[u]&1)); 100 } 101 } 102 } 103 return 0; 104 }
树状数组
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn = 200010; 4 vector<int>g[maxn]; 5 int c[2][maxn],val[maxn],d[maxn],L[maxn],R[maxn],clk,n,m; 6 void dfs(int u,int fa,int dep){ 7 d[u] = dep; 8 L[u] = ++clk; 9 for(int i = g[u].size()-1; i >= 0; --i){ 10 if(g[u][i] == fa) continue; 11 dfs(g[u][i],u,dep+1); 12 } 13 R[u] = clk; 14 } 15 void update(int cur,int i,int val){ 16 while(i < maxn){ 17 c[cur][i] += val; 18 i += i&-i; 19 } 20 } 21 int sum(int cur,int i,int ret = 0){ 22 while(i > 0){ 23 ret += c[cur][i]; 24 i -= i&-i; 25 } 26 return ret; 27 } 28 int main(){ 29 int u,v,op,x; 30 while(~scanf("%d%d",&n,&m)){ 31 for(int i = 1; i <= n; ++i){ 32 g[i].clear(); 33 scanf("%d",val+i); 34 } 35 for(int i = 1; i < n; ++i){ 36 scanf("%d%d",&u,&v); 37 g[u].push_back(v); 38 g[v].push_back(u); 39 } 40 dfs(1,-1,0); 41 memset(c,0,sizeof c); 42 while(m--){ 43 scanf("%d%d",&op,&x); 44 if(op == 1){ 45 scanf("%d",&v); 46 update(d[x]&1,L[x],v); 47 update(d[x]&1,R[x]+1,-v); 48 }else printf("%d\n",val[x] + sum(d[x]&1,L[x]) - sum(~d[x]&1,L[x])); 49 } 50 } 51 return 0; 52 }
