Alyona and a tree CodeForces - 739B (线段树合并)
作者:互联网
大意: 给定有根树, 每个点$x$有权值$a_x$, 对于每个点$x$, 求出$x$子树内所有点$y$, 需要满足$dist(x,y)<=a_y$.
刚开始想错了, 直接打线段树合并了.....因为范围是$long \space long$常数极大, 空间很可能会被卡, 不过竟然过了. 实际上本题每个点对树链上的贡献是单调的, 直接二分就行了
放一下线段树合并代码
#include <iostream> #include <algorithm> #include <cstdio> #include <math.h> #include <set> #include <map> #include <queue> #include <string> #include <string.h> #include <bitset> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc tr[o].l #define rc tr[o].r #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl '\n' using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} //head const int N = 2e5+10; int n, tot, a[N], rt[N], ans[N]; struct _ {int to,w;}; vector<_> g[N]; ll d[N], L, R; struct {int l,r,sum;} tr[N<<6]; typedef int&& dd; int merge(int x, int y) { if (!x||!y) return x+y; tr[x].l=merge(tr[x].l,tr[y].l); tr[x].r=merge(tr[x].r,tr[y].r); tr[x].sum+=tr[y].sum; return x; } int query(int o, ll l, ll r, ll ql, ll qr) { if (!o||ql<=l&&r<=qr) return tr[o].sum; int ans = 0; if (mid>=ql) ans+=query(ls,ql,qr); if (mid<qr) ans+=query(rs,ql,qr); return ans; } void update(int &o, ll l, ll r, ll x) { if (!o) o=++tot; ++tr[o].sum; if (l==r) return; if (mid>=x) update(ls,x); else update(rs,x); } void dfs(int x) { for (auto &&e:g[x]) d[e.to]=d[x]+e.w,dfs(e.to); L = min(L, d[x]), R = max(R, d[x]); L = min(L, d[x]-a[x]), R = max(R, d[x]-a[x]); } void solve(int x) { for (auto &&e:g[x]) { solve(e.to); rt[x]=merge(rt[x],rt[e.to]); } ans[x] = query(rt[x],L,R,L,d[x]); update(rt[x],L,R,d[x]-a[x]); } int main() { int &&t = n; scanf("%d", &n); REP(i,1,n) scanf("%d", a+i); REP(i,2,n) { int f, w; scanf("%d%d", &f, &w); g[f].pb({i,w}); } dfs(1),solve(1); REP(i,1,n) printf("%d ", ans[i]);hr; }
标签:rt,Alyona,int,ll,tree,CodeForces,include,REP,define 来源: https://www.cnblogs.com/uid001/p/10597609.html