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

题目链接：

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

D. Alyona and a tree

time limit per test2 seconds
memory 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-----------------------------------------------------------------------