QTREE3 spoj 2798. Query on a tree again! 树链剖分+线段树

Query on a tree again!

 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;

}