QTREE 树链剖分---模板 spoj QTREE
《树链剖分及其应用》 一文讲得非常清楚,我一早上就把他学会了并且A了这题的入门题。
spoj QTREE
题目:
给出一棵树,有两种操作:
1.修改一条边的边权。
2.询问节点a到b的最大边权。
直接粘代码。更成熟的代码可以看下一篇BZOJ 1036: [ZJOI2008]树的统计Count
#include <set> #include <map> #include <list> #include <cmath> #include <queue> #include <stack> #include <string> #include <vector> #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; typedef long long ll; typedef unsigned long long ull; #define debug puts("here") #define rep(i,n) for(int i=0;i<n;i++) #define rep1(i,n) for(int i=1;i<=n;i++) #define REP(i,a,b) for(int i=a;i<=b;i++) #define foreach(i,vec) for(unsigned i=0;i<vec.size();i++) #define pb push_back #define RD(n) scanf("%d",&n) #define RD2(x,y) scanf("%d%d",&x,&y) #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define RD4(x,y,z,w) scanf("%d%d%d%d",&x,&y,&z,&w) #define All(vec) vec.begin(),vec.end() #define MP make_pair #define PII pair<int,int> #define PQ priority_queue #define cmax(x,y) x = max(x,y) #define cmin(x,y) x = min(x,y) #define Clear(x) memset(x,0,sizeof(x)) /* #pragma comment(linker, "/STACK:1024000000,1024000000") int size = 256 << 20; // 256MB char *p = (char*)malloc(size) + size; __asm__("movl %0, %%esp\n" :: "r"(p) ); */ /******** program ********************/ const int MAXN = 100005; int val[MAXN]; int po[MAXN],tol; bool use[MAXN]; int n; struct node{ int y,c,next; }edge[MAXN]; inline void add(int x,int y,int c){ edge[++tol].y = y; edge[tol].c = c; edge[tol].next = po[x]; po[x] = tol; } struct tc{ // tree chain subdivision int sz; // x子树大小 int dep; // 节点x的深度 int top; // 节点x所在链的顶端节点 int fa; // 节点x的父亲 int son; // 重儿子 int tid; // 在线段树中的位置 }p[MAXN]; void dfsFind(int x,int fa,int dep){ use[x] = true; p[x].dep = dep; p[x].sz = 1; p[x].son = 0; p[x].fa = fa; int mx = 0; for(int i=po[x];i;i=edge[i].next){ int y = edge[i].y; if(use[y])continue; dfsFind(y,x,dep+1); p[x].sz += p[y].sz; if(p[y].sz>mx){ p[x].son = y; mx = p[y].sz; } } } int tid; void dfsCon(int x,int fa){ use[x] = true; p[x].tid = ++ tid; p[x].top = fa; if(p[x].son) dfsCon(p[x].son,fa); for(int i=po[x];i;i=edge[i].next){ int y = edge[i].y; if(use[y])continue; dfsCon(y,y); } } struct Tree{ int l,r,mx; inline int mid(){ return (l+r)>>1; } }tree[MAXN<<2]; inline void update(int rt){ tree[rt].mx = max(tree[rt<<1].mx,tree[rt<<1|1].mx); } void build(int l,int r,int rt){ tree[rt].l = l; tree[rt].r = r; if(l==r){ tree[rt].mx = val[l]; return; } int mid = tree[rt].mid(); build(l,mid,rt<<1); build(mid+1,r,rt<<1|1); update(rt); } void modify(int pos,int c,int rt){ if(tree[rt].l==tree[rt].r){ tree[rt].mx = c; return; } int mid = tree[rt].mid(); if(pos<=mid) modify(pos,c,rt<<1); else modify(pos,c,rt<<1|1); update(rt); } int ask(int l,int r,int rt){ if(tree[rt].l==l&&tree[rt].r==r) return tree[rt].mx; int mid = tree[rt].mid(); if(r<=mid) return ask(l,r,rt<<1); else if(l>mid) return ask(l,r,rt<<1|1); else return max( ask(l,mid,rt<<1),ask(mid+1,r,rt<<1|1) ); } int main(){ #ifndef ONLINE_JUDGE freopen("sum.in","r",stdin); //freopen("sum.out","w",stdout); #endif int x,y,z,ncase; RD(ncase); while(ncase--){ RD(n); Clear(po); tol = 1; REP(i,2,n){ RD3(x,y,z); add(x,y,z); add(y,x,z); } Clear(use); dfsFind(1,1,1); tid = 0; Clear(use); dfsCon(1,1); for(int i=2;i<tol;i+=2){ int x = edge[i^1].y; // 对应于第x条边的节点 (x,y) int y = edge[i].y; if(p[x].dep>p[y].dep) val[ p[x].tid ] = edge[i].c; else val[ p[y].tid ] = edge[i].c; } build(2,n,1); char op[10]; while(scanf("%s",op),op[0]!='D'){ if(op[0]=='C'){ RD2(x,z); y = edge[x<<1].y; x = edge[x<<1|1].y; if( p[x].dep>p[y].dep ) modify( p[x].tid,z,1 ); else modify( p[y].tid,z,1 ); }else{ RD2(x,y); int ans = -(1<<30); while( p[x].top != p[y].top ){ if( p[ p[x].top ].dep < p[ p[y].top ].dep ) swap(x,y); ans = max(ans,ask(p[ p[x].top ].tid,p[x].tid,1)); x = p[ p[x].top ].fa; } if(p[x].dep>p[y].dep) swap(x,y); if(x!=y) ans = max(ans,ask(p[x].tid+1,p[y].tid,1)); printf("%d\n",ans); } } } return 0; }