Big binary tree
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 670 Accepted Submission(s): 235
Problem Description
You are given a complete binary tree with n nodes. The root node is numbered 1, and node x's father node is ⌊x/2⌋. At the beginning, node x has a value of exactly x. We define the value of a path as the sum of all nodes it passes(including two ends, or one if the path only has one node). Now there are two kinds of operations:
1. change u x Set node u's value as x(1≤u≤n;1≤x≤10^10)
2. query u Query the max value of all paths which passes node u.
Input
There are multiple cases.
For each case:
The first line contains two integers n,m(1≤n≤10^8,1≤m≤10^5), which represent the size of the tree and the number of operations, respectively.
Then m lines follows. Each line is an operation with syntax described above.
For each case:
The first line contains two integers n,m(1≤n≤10^8,1≤m≤10^5), which represent the size of the tree and the number of operations, respectively.
Then m lines follows. Each line is an operation with syntax described above.
Output
For each query operation, output an integer in one line, indicating the max value of all paths which passes the specific node.
Sample Input
6 13
query 1
query 2
query 3
query 4
query 5
query 6
change 6 1
query 1
query 2
query 3
query 4
query 5
query 6
Sample Output
17
17
17
16
17
17
12
12
12
11
12
12
Source
Recommend
题意：题目是给一棵完全二叉树，从上到下从左到右给每个节点标号，每个点有权值，初始权值为其标号，然后有两种操作：
1、把u点权值改为x
2、查询所有经过u点的路径中，路径上的点权和最大。
1、把u点权值改为x
2、查询所有经过u点的路径中，路径上的点权和最大。
思路：每次修改只会改变log(n)个节点，最多改变log(n)*m个节点，约为2.7*10^6个节点，但是n最大为10^8，所以只能用map存储。没有修改过的结点不需要存储，因为没有修改过的话一定是右儿子大，有以只需要往右走就能寻找到以当前点为根的最大路径点权和，但是如果左边的结点数比右边的多的话就需要往左了。
代码：
View Code
#include<iostream> #include<cstdio> #include<cmath> #include<cstring> #include<algorithm> #include<set> #include<bitset> #include<map> #include<queue> #include<stack> #include<vector> using namespace std; typedef long long ll; typedef pair<int,int> P; #define bug(x) cout<<"bug"<<x<<endl; #define PI acos(-1.0) #define eps 1e-8 const int N=2e5+100,M=1e5+100; const int inf=0x3f3f3f3f; const ll INF=1e18+7,mod=1e9+7; ll n; char s[100]; map<ll,ll> vis,deep; ll getnum(ll tmp) { ll ans=0; while(tmp<=n) { if(deep.find(tmp)!=deep.end()) return ans+deep[tmp]; if(vis.find(tmp)==vis.end()) ans+=tmp; else ans+=vis[tmp]; ll tmp1=tmp<<1,tmp2=tmp<<1|1; int h1=0,h2=0; while(tmp1<=n) tmp1=tmp1<<1,h1++; while(tmp2<=n) tmp2=tmp2<<1,h2++; if(h1==h2) tmp=tmp<<1|1; else tmp=tmp<<1; } return ans; } void getchild(ll u,ll &ch1,ll &ch2) { ch1=getnum(u<<1),ch2=getnum(u<<1|1); } int main() { int m; while(scanf("%lld%d",&n,&m)!=EOF) { vis.clear(),deep.clear(); for(int i=1; i<=m; i++) { ll u,x; scanf("%s %lld",s,&u); if(s[0]=='c') { scanf("%lld",&x); vis[u]=x; while(u>=1) { ll ch1,ch2; getchild(u,ch1,ch2); if(vis.find(u)==vis.end()) deep[u]=max(ch1,ch2)+u; else deep[u]=max(ch1,ch2)+vis[u]; u/=2; } } else { ll ch1,ch2; getchild(u,ch1,ch2); u=ch1>ch2?(u<<1):(u<<1|1); ll cou=max(ch1,ch2); ll ans=0; while(u>1) { if(vis.find(u/2)==vis.end()) cou+=u/2; else cou+=vis[u/2]; ans=max(ans,getnum(u^1)+cou); u/=2; } printf("%lld ",ans); } } } return 0; }