给出一棵树,树节点的颜色初始时为白色,有两种操作:
0.把节点x的颜色置反(黑变白,白变黑)。
1.询问节点1到节点x的路径上第一个黑色节点的编号。
分析:
先树链剖分,线段树节点维护深度最浅的节点编号。
注意到,如果以节点1为树根时,显然每条重链在一个区间,并且区间的左端会出现在深度浅的地方。所以每次查找时发现左区间有的话,直接更新答案。
9929151 | 2013-08-28 10:45:55 | Query on a tree again! | 100 edit run |
12.54 | 27M |
C++ 4.3.2 |
#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 " :: "r"(p) ); */ /******** program ********************/ const int MAXN = 200005; int son[MAXN],tid[MAXN],top[MAXN],dep[MAXN],fa[MAXN],sz[MAXN],tim; bool use[MAXN]; int id[MAXN]; int po[MAXN],tol; struct segTree{ int l,r,pos,c; inline int mid(){ return (l+r)>>1; } }tree[MAXN<<2]; struct Edge{ int y,next; }edge[MAXN<<1]; inline void add(int x,int y){ edge[++tol].y = y; edge[tol].next = po[x]; po[x] = tol; } // 树链剖分部分 void dfsFind(int x,int pa,int depth){ dep[x] = depth; fa[x] = pa; sz[x] = 1; son[x] = 0; for(int i=po[x];i;i=edge[i].next){ int y = edge[i].y; if(y==pa)continue; dfsFind(y,x,depth+1); sz[x] += sz[y]; if(sz[y]>sz[ son[x] ]) son[x] = y; } } void dfsCon(int x,int pa){ use[x] = true; top[x] = pa; tid[x] = ++ tim; if(son[x])dfsCon(son[x],pa); for(int i=po[x];i;i=edge[i].next){ int y = edge[i].y; if(use[y])continue; dfsCon(y,y); } } void build(int l,int r,int rt){ tree[rt].l = l; tree[rt].r = r; tree[rt].pos = 0; tree[rt].c = 0; if(l==r) return; int mid = tree[rt].mid(); build(l,mid,rt<<1); build(mid+1,r,rt<<1|1); } void modify(int pos,int rt){ if(tree[rt].l==tree[rt].r){ tree[rt].c ^= 1; if(tree[rt].c) tree[rt].pos = id[tree[rt].l]; else tree[rt].pos = 0; return; } int mid = tree[rt].mid(); if(pos<=mid)modify(pos,rt<<1); else modify(pos,rt<<1|1); if(tree[rt<<1].c){ tree[rt].c = tree[rt<<1].c; tree[rt].pos = tree[rt<<1].pos; }else{ tree[rt].c = tree[rt<<1|1].c; tree[rt].pos = tree[rt<<1|1].pos; } } int ask(int l,int r,int rt){ if(tree[rt].c==0)return 0; if(l<=tree[rt].l&&tree[rt].r<=r) return tree[rt].pos; 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{ int t = ask(l,r,rt<<1); if(t)return t; return ask(l,r,rt<<1|1); } } inline int ask(int y){ int x = 1; int ans = -1; while(top[x]!=top[y]){ if(dep[top[x]]<dep[top[y]]) swap(x,y); int t = ask(tid[top[x]],tid[x],1); if(t)ans = t; x = fa[top[x]]; } if(dep[x]>dep[y])swap(x,y); int t = ask(tid[x],tid[y],1); if(t)ans = t; return ans; } int main(){ #ifndef ONLINE_JUDGE freopen("sum.in","r",stdin); //freopen("sum.out","w",stdout); #endif int x,y,n,q,op; while(~RD2(n,q)){ Clear(po); tol = 0; REP(i,2,n){ RD2(x,y); add(x,y); add(y,x); } dfsFind(1,1,1); tim = 0; Clear(use); dfsCon(1,1); rep1(i,n) id[ tid[i] ] = i; build(1,n,1); while(q--){ RD2(op,x); if(op==0) modify(tid[x],1); else printf("%d ",ask(x)); } } return 0; }