HDU 6161.Big binary tree 二叉树
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

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.
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.
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。
#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+,M=1e5+;
const int inf=0x3f3f3f3f;
const ll INF=1e18+,mod=1e9+;
ll n;
char s[];
map<ll,ll> vis,deep;
ll getnum(ll tmp)
{
ll ans=;
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<<,tmp2=tmp<<|;
int h1=,h2=;
while(tmp1<=n) tmp1=tmp1<<,h1++;
while(tmp2<=n) tmp2=tmp2<<,h2++;
if(h1==h2) tmp=tmp<<|;
else tmp=tmp<<;
}
return ans;
}
void getchild(ll u,ll &ch1,ll &ch2)
{
ch1=getnum(u<<),ch2=getnum(u<<|);
}
int main()
{
int m;
while(scanf("%lld%d",&n,&m)!=EOF)
{
vis.clear(),deep.clear();
for(int i=; i<=m; i++)
{
ll u,x;
scanf("%s %lld",s,&u);
if(s[]=='c')
{
scanf("%lld",&x);
vis[u]=x;
while(u>=)
{
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/=;
}
}
else
{
ll ch1,ch2;
getchild(u,ch1,ch2);
u=ch1>ch2?(u<<):(u<<|);
ll cou=max(ch1,ch2);
ll ans=;
while(u>)
{
if(vis.find(u/)==vis.end()) cou+=u/;
else cou+=vis[u/];
ans=max(ans,getnum(u^)+cou);
u/=;
}
printf("%lld\n",ans);
}
}
}
return ;
}
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