QTREE3 spoj 2798. Query on a tree again! 树链剖分+线段树
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\n" :: "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\n",ask(x));
}
}
return 0;
}
