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Codeforces Round #381 (Div. 2) D. Alyona and a tree 树上二分+前缀和思想

2016-12-06 16:29 232 查看

题目链接：

http://codeforces.com/contest/740/problem/D

D. Alyona and a tree

time limit per test2 secondsmemory limit per test256 megabytes

问题描述


Alyona has a tree with n vertices. The root of the tree is the vertex 1. In each vertex Alyona wrote an positive integer, in the vertex i she wrote ai. Moreover, the girl wrote a positive integer to every edge of the tree (possibly, different integers on different edges).

Let's define dist(v, u) as the sum of the integers written on the edges of the simple path from v to u.

The vertex v controls the vertex u (v ≠ u) if and only if u is in the subtree of v and dist(v, u) ≤ au.

Alyona wants to settle in some vertex. In order to do this, she wants to know for each vertex v what is the number of vertices u such that v controls u.


输入


The first line contains single integer n (1 ≤ n ≤ 2·105).

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the integers written in the vertices.

The next (n - 1) lines contain two integers each. The i-th of these lines contains integers pi and wi (1 ≤ pi ≤ n, 1 ≤ wi ≤ 109) — the parent of the (i + 1)-th vertex in the tree and the number written on the edge between pi and (i + 1).

It is guaranteed that the given graph is a tree.


输出


Print n integers — the i-th of these numbers should be equal to the number of vertices that the i-th vertex controls


样例输入


5

2 5 1 4 6

1 7

1 1

3 5

3 6


样例输出


1 0 1 0 0


题意


给你一颗点权为a[i],的带边权的有根树(树根为1),对于节点v和它的子节点u之间，我们称v控制了u当且仅当dis(v,u)<=a[u]的时候，现在让你求每个结点能控制的子节点的个数。


题解

二分+前缀和。

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef __int64 LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=2e5+10;
int n;
int arr[maxn];

VPII G[maxn];
LL dep[maxn];
///ans维护的是前缀和，这里的前缀指的是从叶子到根的方向
int ans[maxn];

///path维护当前的一条路径
vector<pair<LL,int> > path;
void dfs(int u){
///自己肯定能够的到自己
ans[u]++;
///二分找第一个dis(ancestor,u)>arr[u]既dep[u]-dep[ancestor]>arr[u]既dep[u]-arr[u]>dep[ancesotor];
int p=lower_bound(all(path),mkp(dep[u]-arr[u],-1))-path.begin()-1;
if(p>=0) ans[path[p].Y]--;

path.pb(mkp(dep[u],u));
for(int i=0;i<G[u].sz();i++){
int v=G[u][i].X;
dep[v]=dep[u]+G[u][i].Y;
dfs(v);
ans[u]+=ans[v];
}
path.pop_back();
}

int main(){
clr(ans,0);
scf("%d",&n);
for(int i=1;i<=n;i++) scf("%d",&arr[i]);
for(int v=2;v<=n;v++){
int u,w;
scf("%d%d",&u,&w);
G[u].pb(mkp(v,w));
}

dep[1]=0;
dfs(1);

for(int i=1;i<=n;i++){
prf("%d",ans[i]-1);
if(i==n) prf("\n");
else prf(" ");
}

return 0;
}

//end----------------------------------------------------------------------


树上倍增+前缀和


#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef __int64 LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=2e5+10;
const int maxm=22;
int n;
int arr[maxn];
VPII G[maxn];

///anc[i][j]表示i节点的2^j的祖先
int anc[maxn][maxm];
int ans[maxn];
LL dep[maxn];
void dfs(int u,int f){
ans[u]++;
anc[u][0]=f;
for(int i=1;i<maxm;i++){
anc[u][i]=anc[anc[u][i-1]][i-1];
}

///树上倍增
int pos=u;
for(int i=maxm-1;i>=0;i--){
int tmp=anc[pos][i];
if(dep[u]-dep[tmp]<=arr[u]){
pos=tmp;
}
}

pos=anc[pos][0];
ans[pos]--;

for(int i=0;i<G[u].sz();i++){
int v=G[u][i].X;
dep[v]=dep[u]+G[u][i].Y;
dfs(v,u);
ans[u]+=ans[v];
}
}

int main(){
clr(ans,0);
scf("%d",&n);
for(int i=1;i<=n;i++) scf("%d",&arr[i]);
for(int v=2;v<=n;v++){
int u,w;
scf("%d%d",&u,&w);
G[u].pb(mkp(v,w));
}

dep[1]=0;
dfs(1,0);

for(int i=1;i<=n;i++){
prf("%d",ans[i]-1);
if(i==n) prf("\n");
else prf(" ");
}

return 0;
}

//end-----------------------------------------------------------------------
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