AC日记——Propagating tree Codeforces 383c
2017-03-14 21:28
351 查看
C. Propagating tree
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Iahub likes trees very much. Recently he discovered an interesting tree named propagating tree. The tree consists of n nodes numbered from 1 to n, each node i having an initial value ai. The root of the tree is node 1.
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 vi and 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.
Examples
input
output
Note
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)
思路;
dfs序同时处理深度,然后搞搞就a了;
来,上代码:
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Iahub likes trees very much. Recently he discovered an interesting tree named propagating tree. The tree consists of n nodes numbered from 1 to n, each node i having an initial value ai. The root of the tree is node 1.
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 vi and 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.
Examples
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
output
3 3 0
Note
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)
思路;
dfs序同时处理深度,然后搞搞就a了;
来,上代码:
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define maxn 200005 #define LL long long using namespace std; struct EdgeType { int v,e; }; struct EdgeType edge[maxn<<1]; struct TreeNodeType { int l,r,dis,mid,flag; }; struct TreeNodeType tree[maxn<<2]; int if_z,n,m,dis[maxn],head[maxn]; int cnt,f[maxn],li[maxn],ri[maxn]; int id[maxn],dis_[maxn],type,x,ans; bool deep[maxn]; char Cget; inline void in(int &now) { now=0,if_z=1,Cget=getchar(); while(Cget>'9'||Cget<'0') { if(Cget=='-') if_z=-1; Cget=getchar(); } while(Cget>='0'&&Cget<='9') { now=now*10+Cget-'0'; Cget=getchar(); } now*=if_z; } void search_1(int now,int fa) { deep[now]=!deep[fa],f[now]=fa; id[now]=++cnt,li[now]=cnt,dis_[cnt]=dis[now]; if(!deep[now]) dis_[cnt]*=-1; for(int i=head[now];i;i=edge[i].e) { if(edge[i].v==f[now]) continue; search_1(edge[i].v,now); } ri[now]=cnt; } void tree_build(int now,int l,int r) { tree[now].l=l,tree[now].r=r; if(l==r) { tree[now].dis=dis_[l]; return ; } tree[now].mid=(l+r)>>1; tree_build(now<<1,l,tree[now].mid); tree_build(now<<1|1,tree[now].mid+1,r); } void tree_do(int now,int l,int r) { if(tree[now].l==l&&tree[now].r==r) { if(type==1) { tree[now].flag+=x; if(l==r) tree[now].dis+=x; } else ans+=tree[now].dis; return ; } if(tree[now].flag) { tree[now<<1].flag+=tree[now].flag; tree[now<<1|1].flag+=tree[now].flag; if(tree[now<<1].l==tree[now<<1].r) tree[now<<1].dis+=tree[now].flag; if(tree[now<<1|1].l==tree[now<<1|1].r) tree[now<<1|1].dis+=tree[now].flag; tree[now].flag=0; } if(l>tree[now].mid) tree_do(now<<1|1,l,r); else if(r<=tree[now].mid) tree_do(now<<1,l,r); else { tree_do(now<<1,l,tree[now].mid); tree_do(now<<1|1,tree[now].mid+1,r); } } int main() { in(n),in(m);int u,v; for(int i=1;i<=n;i++) in(dis[i]); for(int i=1;i<n;i++) { in(u),in(v); edge[++cnt].v=v,edge[cnt].e=head[u],head[u]=cnt; edge[++cnt].v=u,edge[cnt].e=head[v],head[v]=cnt; } cnt=0,search_1(1,0); tree_build(1,1,n); while(m--) { in(type),in(u); if(type==1) { in(x); if(!deep[u]) x*=-1; tree_do(1,li[u],ri[u]); } else { ans=0; tree_do(1,li[u],li[u]); if(!deep[u]) ans*=-1; printf("%d\n",ans); } } return 0; }
相关文章推荐
- AC日记——Is it rated? codeforces 807a
- AC日记——Dishonest Sellers Codeforces 779c
- AC日记——Weird Rounding Codeforces 779b
- AC日记——Mice and Holes codeforces 797f
- AC日记——Success Rate codeforces 807c
- AC日记——Valued Keys codeforces 801B
- AC日记——Broken BST codeforces 797d
- AC日记——Dynamic Problem Scoring codeforces 807d
- AC日记——Cards Sorting codeforces 830B
- AC日记——Array Queries codeforces 797e
- AC日记——Sagheer and Crossroads codeforces 812a
- AC日记——黑魔法师之门 codevs 1995
- AC日记——联合权值 洛谷 P1351
- AC日记——旅行 洛谷 P3313
- AC日记——最大子树和 洛谷 P1122
- AC日记——最小路径覆盖问题 洛谷 P2764
- AC日记——[USACO15DEC]最大流Max Flow 洛谷 P3128
- AC日记——[USACO09JAN]全流Total Flow 洛谷 P2936
- AC日记——[USACO09OCT]Bessie的体重问题Bessie's We… 洛谷 P2639
- AC日记——NASA的食物计划 洛谷 P1507