codeforces contest 383 problem C(树状数组+dfs序)
2017-10-11 10:46
585 查看
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
题意:给定一棵树,每次进行两种操作,将u节点的权值增加v,他的儿子节点减v,以此递归;问u节点的值为多少;
解:dfs标上序号,开两个数组一个记录加权,一个记录减权
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6+10;
typedef long long LL;
vector<int>p
;
struct node
{
int l, r, v, d;
}g
;
int cnt;
void dfs(int u,int f,int d)
{
g[u].l=++cnt,g[u].d=d;
for(int i=0;i<p[u].size();i++)
{
int v=p[u][i];
if(v==f) continue;
dfs(v,u,1-d);
}
g[u].r=++cnt;
return ;
}
int c[2]
;
void add(int x,int v,int cx[])
{
while(x<=cnt)
{
cx[x]+=v;
x+=(x&(-x));
}
return ;
}
int get(int u,int cx[])
{
int sum=0;
while(u)
{
sum+=cx[u];
u-=(u&(-u));
}
return sum;
}
int main()
{
int n, m;
scanf("%d %d", &n, &m);
for(int i=1;i<=n;i++) scanf("%d",&g[i].v);
for(int i=1;i<n;i++)
{
int x, y;
scanf("%d %d", &x, &y);
p[x].push_back(y),p[y].push_back(x);
}
cnt=0;
dfs(1,-1,1);
while(m--)
{
int x, u, v;
scanf("%d", &x);
if(x==1)
{
scanf("%d %d",&u,&v);
add(g[u].l,v,c[g[u].d]),add(g[u].r+1,-v,c[g[u].d]);
}
else
{
scanf("%d",&u);
printf("%d\n",g[u].v+get(g[u].l,c[g[u].d])-get(g[u].l,c[1-g[u].d]));
}
}
return 0;
}
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
题意:给定一棵树,每次进行两种操作,将u节点的权值增加v,他的儿子节点减v,以此递归;问u节点的值为多少;
解:dfs标上序号,开两个数组一个记录加权,一个记录减权
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6+10;
typedef long long LL;
vector<int>p
;
struct node
{
int l, r, v, d;
}g
;
int cnt;
void dfs(int u,int f,int d)
{
g[u].l=++cnt,g[u].d=d;
for(int i=0;i<p[u].size();i++)
{
int v=p[u][i];
if(v==f) continue;
dfs(v,u,1-d);
}
g[u].r=++cnt;
return ;
}
int c[2]
;
void add(int x,int v,int cx[])
{
while(x<=cnt)
{
cx[x]+=v;
x+=(x&(-x));
}
return ;
}
int get(int u,int cx[])
{
int sum=0;
while(u)
{
sum+=cx[u];
u-=(u&(-u));
}
return sum;
}
int main()
{
int n, m;
scanf("%d %d", &n, &m);
for(int i=1;i<=n;i++) scanf("%d",&g[i].v);
for(int i=1;i<n;i++)
{
int x, y;
scanf("%d %d", &x, &y);
p[x].push_back(y),p[y].push_back(x);
}
cnt=0;
dfs(1,-1,1);
while(m--)
{
int x, u, v;
scanf("%d", &x);
if(x==1)
{
scanf("%d %d",&u,&v);
add(g[u].l,v,c[g[u].d]),add(g[u].r+1,-v,c[g[u].d]);
}
else
{
scanf("%d",&u);
printf("%d\n",g[u].v+get(g[u].l,c[g[u].d])-get(g[u].l,c[1-g[u].d]));
}
}
return 0;
}
相关文章推荐
- 【codeforces】2014 Asia Xian Regional Contest G The Problem to Slow Down You 【Palindromic Tree】
- Codeforces contest 883 problem L. Berland.Taxi(Treap+优先队列)
- codeforces contest 551 problem E(分块)
- codeforces contest 13 problem E(分块)
- codeforces contest 869 problem C(组合数)
- codeforces contest 855 problem B(前缀后缀)
- codeforces contest 855 problem C(树形DP)
- codeforces contest 868 problem D(玄学)
- The 7th Zhejiang Provincial Collegiate Programming Contest->Problem B:B - Somali Pirates
- Codeforces Round #439 (Div. 2) Problem E (Codeforces 869E) - 暴力 - 随机化 - 二维树状数组 - 差分
- 【Codeforces】2015-2016 ACM-ICPC Nordic Collegiate Programming Contest (NCPC 2015) A Adjoin the Netwo
- CodeForces - 633B A Trivial Problem(找规律)
- 2015 Multi-University Training Contest 7 hdu 5373 The shortest problem
- The 6th Zhejiang Provincial Collegiate Programming Contest->Problem I:A Stack or A Queue?
- Problem - 218B - Codeforces
- 【codeforces 749A】Bachgold Problem
- 【codeforces】 Far Relative’s Problem
- Codeforces 2016 ACM Amman Collegiate Programming Contest B. The Little Match Girl(贪心)
- codeforces contest 779 B题
- codeforces 165-C. Another Problem on Strings(计数+尺取)