HDU4718 树链剖分

xiaoxiao2021-02-27  540

题目:HDU4718

我正在强行骗自己会树剖了 原来的blog树剖写的有点错误(在dfs1求son那里),现已改正 给出一棵树和树上权值,求从u到v的最长连续上升子序列长度 很坑爹。。线段树上维护了8个值。。维护上升下降的max值、最左最右节点的值和向左向右的上升下降长度 常数很大很大很大。。。。 这一坨东西改的我都要疯了。。细节写错了一个地方,结果对拍debug了三个小时,真是作死

#include <cstdio> #include <iostream> #include <algorithm> #include <vector> #include <set> #include <map> #include <string> #include <stack> #include <queue> #include <utility> using namespace std; #define REP(I,N) for (I=0;I<N;I++) #define rREP(I,N) for (I=N-1;I>=0;I--) #define rep(I,S,N) for (I=S;I<N;I++) #define rrep(I,S,N) for (I=N-1;I>=S;I--) #define FOR(I,S,N) for (I=S;I<=N;I++) #define mp(A,B) make_pair(A,B) typedef unsigned long long ULL; typedef long long LL; const int INF=0x3f3f3f3f; const LL INFF=0x3f3f3f3f3f3f3f3fll; const LL hash=1e9+7; const LL maxn=1e5+7; const double eps=0.00000001; LL gcd(LL a,LL b){return b?gcd(b,a%b):a;} template<typename T>inline T abs(T a,T b) {return a>0?a:-a;} int tot; struct node{ int lval,rval,ldown,lup,rdown,rup,upmx,downmx; node():upmx(0),downmx(0){}; }tree[maxn<<2]; int a[maxn]; node merge(node L,node R){ if (L.upmx==0) return R; if (R.upmx==0) return L; node ret; ret.upmx=max(L.upmx,R.upmx); ret.downmx=max(L.downmx,R.downmx); ret.lval=L.lval; ret.lup=L.lup; ret.ldown=L.ldown; ret.rval=R.rval; ret.rup=R.rup; ret.rdown=R.rdown; if (L.rval<R.lval){ ret.upmx=max(ret.upmx,L.rup+R.lup); if (L.downmx==1) ret.lup=L.lup+R.lup; if (R.downmx==1) ret.rup=L.rup+R.rup; } if (L.rval>R.lval){ ret.downmx=max(ret.downmx,L.rdown+R.ldown); if (L.upmx==1) ret.ldown=L.ldown+R.ldown; if (R.upmx==1) ret.rdown=L.rdown+R.rdown; } return ret; } void build(int x,int l,int r){ if (l==r){ tree[x].lval=tree[x].rval=a[l]; tree[x].lup=tree[x].ldown=tree[x].rup=tree[x].rdown=tree[x].upmx=tree[x].downmx=1; return; } int mid=(l+r)/2; build(x<<1,l,mid); build(x<<1|1,mid+1,r); tree[x]=merge(tree[x<<1],tree[x<<1|1]); } node query(int x,int l,int r,int L,int R){ node ret; if (l<=L&&R<=r) return tree[x]; int mid=(L+R)/2; if (mid>=l&&r>mid) return merge(query(x<<1,l,r,L,mid),query(x<<1|1,l,r,mid+1,R)); if (mid>=l) return query(x<<1,l,r,L,mid); return query(x<<1|1,l,r,mid+1,R); } int n,i,j,q; int u,v; vector<int> edge[maxn]; int fa[maxn],son[maxn],top[maxn],dep[maxn],id[maxn],sz[maxn]; int b[maxn]; void dfs1(int u,int depth){ int v,i,mx=-1; son[u]=0;sz[u]=1;dep[u]=depth; REP(i,edge[u].size()){ v=edge[u][i]; dfs1(v,depth+1); sz[u]+=sz[v]; if (sz[v]>mx) mx=sz[v],son[u]=v; } } void dfs2(int u,int x){ int v,i; top[u]=x;id[u]=++tot; if (son[u]) dfs2(son[u],x); REP(i,edge[u].size()){ v=edge[u][i]; if (v==fa[u]||v==son[u]) continue; dfs2(v,v); } } int Query(int x,int y){ node up,down; int ret,mark1=0,mark2=0; while (top[x]!=top[y]){ if (dep[top[x]]>dep[top[y]]){ up=merge(query(1,id[top[x]],id[x],1,tot),up); x=fa[top[x]]; mark1=1; }else { down=merge(query(1,id[top[y]],id[y],1,tot),down); y=fa[top[y]]; mark2=1; } } if (dep[x]>dep[y]) up=merge(query(1,id[y],id[x],1,tot),up),mark1=1; else down=merge(query(1,id[x],id[y],1,tot),down),mark2=1; ret=max(up.downmx,down.upmx); if (mark1&&mark2&&up.lval<down.lval) ret=max(ret,up.ldown+down.lup); return ret; } int T,t; int main(){ scanf("%d",&T); FOR (t,1,T){ scanf("%d",&n); FOR(i,1,n) edge[i].clear();tot=0; FOR(i,1,n) scanf("%d",&b[i]); FOR(i,2,n){scanf("%d",&fa[i]); edge[fa[i]].push_back(i);} dfs1(1,1); dfs2(1,1); FOR(i,1,n) a[id[i]]=b[i]; build(1,1,tot); scanf("%d",&q); printf("Case #%d:\n",t); while (q--){ scanf("%d%d",&u,&v); printf("%d\n",Query(u,v)); } if (t!=T) puts(""); } } /* */
转载请注明原文地址: https://www.6miu.com/read-624.html

最新回复(0)