SPOJ QTREE3 Query on a tree again! 树链剖分

题目：

https://vjudge.net/problem/SPOJ-QTREE3

题意：

给定一棵树，初始树上的每个点都为白色，有以下操作：

• 0 i：把第i个点的颜色取反，白变黑，黑变白
• 1 v：求从点1到点v的路径上第一个颜色为黑的点

思路：

树链剖分，每次优先查询左子区间，再查询右子区间。因为一定是从v往1走，最后一个查询到的值才是答案，所以查到一个值就更新一下答案

#include <bits/stdc++.h>using namespace std;const int N = 100000 + 10;struct edge{    int to, next;} g[N*2];int cnt, head[N];int dep[N], siz[N], son[N], fat[N], id[N], top[N];int num;int rid[N];struct node{    int l, r, sum;} tr[N*4];void init(){    cnt = 0;    memset(head, -1, sizeof head);    num = 0;}void add_edge(int v, int u){    g[cnt].to = u, g[cnt].next = head[v], head[v] = cnt++;}void dfs1(int v, int fa, int d){    dep[v] = d, son[v] = 0, siz[v] = 1, fat[v] = fa;    for(int i = head[v]; i != -1; i = g[i].next)    {        int u = g[i].to;        if(u != fa)        {            dfs1(u, v, d + 1);            siz[v] += siz[u];            if(siz[son[v]] < siz[u]) son[v] = u;        }    }}void dfs2(int v, int tp){    top[v] = tp, id[v] = ++num;    if(son[v]) dfs2(son[v], top[v]);    for(int i = head[v]; i != -1; i = g[i].next)    {        int u = g[i].to;        if(u != fat[v] && u != son[v]) dfs2(u, u);    }}void push_up(int k){    tr[k].sum = tr[k<<1].sum + tr[k<<1|1].sum;}void build(int l, int r, int k){    tr[k].l = l, tr[k].r = r, tr[k].sum = 0;    if(l == r)    {        tr[k].sum = 0;        return;    }    int mid = (l + r) >> 1;    build(l, mid, k << 1);    build(mid + 1, r, k << 1|1);    push_up(k);}void update(int x, int k){    if(tr[k].l == tr[k].r)    {        tr[k].sum ^= 1;        return;    }    int mid = (tr[k].l + tr[k].r) >> 1;    if(x <= mid) update(x, k << 1);    else update(x, k << 1|1);    push_up(k);}int query(int l, int r, int k){    if(tr[k].sum == 0) return 0;    if(tr[k].l == tr[k].r) return tr[k].l;    int mid = (tr[k].l + tr[k].r) >> 1;    int ans = 0;    if(l <= mid) ans = query(l, r, k << 1);    if(ans) return ans;    if(r > mid) ans = query(l, r, k << 1|1);    return ans;}int Query(int v, int u){    int t1 = top[v], t2 = top[u];    int ans = 0, tmp = 0;    while(t1 != t2)    {        if(dep[t1] < dep[t2])            swap(t1, t2), swap(v, u);        tmp = query(id[t1], id[v], 1);        if(tmp) ans = tmp;        v = fat[t1], t1 = top[v];    }    if(dep[v] > dep[u]) swap(v, u);    tmp = query(id[v], id[u], 1);    if(tmp) ans = tmp;    return ans;}int main(){    int n, m, x, y;    while(~ scanf("%d%d", &n, &m))    {        init();        for(int i = 1; i <= n-1; i++)        {            scanf("%d%d", &x, &y);            add_edge(x, y);            add_edge(y, x);        }        dfs1(1, 0, 1);        dfs2(1, 1);        build(1, num, 1);        for(int i = 1; i <= n; i++) rid[id[i]] = i;        for(int i = 1; i <= m; i++)        {            scanf("%d%d", &x, &y);            if(x == 0) update(id[y], 1);            else            {                int ans = Query(x, y);                printf("%d\n", ans ? rid[ans] : -1);            }        }    }    return 0;}

另外一种写法， 只有在线段树部分不太一样

#include <bits/stdc++.h>using namespace std;const int N = 100000 + 10;struct edge{    int to, next;} g[N*2];int cnt, head[N];int dep[N], siz[N], son[N], fat[N], id[N], top[N];int rid[N];int num;struct node{    int l, r, val, idx;} tr[N*4];void init(){    cnt = 0;    memset(head, -1, sizeof head);    num = 0;}void add_edge(int v, int u){    g[cnt].to = u, g[cnt].next = head[v], head[v] = cnt++;}void dfs1(int v, int fa, int d){    dep[v] = d, son[v] = 0, siz[v] = 1, fat[v] = fa;    for(int i = head[v]; i != -1; i = g[i].next)    {        int u = g[i].to;        if(u != fa)        {            dfs1(u, v, d + 1);            siz[v] += siz[u];            if(siz[son[v]] < siz[u]) son[v] = u;        }    }}void dfs2(int v, int tp){    top[v] = tp, id[v] = ++num;    if(son[v]) dfs2(son[v], top[v]);    for(int i = head[v]; i != -1; i = g[i].next)    {        int u = g[i].to;        if(u != fat[v] && u != son[v]) dfs2(u, u);    }}void push_up(int k){    if(tr[k<<1].val) tr[k].val = tr[k<<1].val, tr[k].idx = tr[k<<1].idx;    else tr[k].val = tr[k<<1|1].val, tr[k].idx = tr[k<<1|1].idx;}void build(int l, int r, int k){    tr[k].l = l, tr[k].r = r, tr[k].val = 0;    if(l == r)    {        tr[k].val = 0;        return;    }    int mid = (l + r) >> 1;    build(l, mid, k << 1);    build(mid + 1, r, k << 1|1);    push_up(k);}void update(int x, int k){    if(tr[k].l == tr[k].r)    {        tr[k].val ^= 1;        if(tr[k].val) tr[k].idx = rid[tr[k].l];        else tr[k].idx = 0;        return;    }    int mid = (tr[k].l + tr[k].r) >> 1;    if(x <= mid) update(x, k << 1);    else update(x, k << 1|1);    push_up(k);}int query(int l, int r, int k){    if(tr[k].val == 0) return 0;    if(l <= tr[k].l && tr[k].r <= r) return tr[k].idx;    int mid = (tr[k].l + tr[k].r) >> 1;    int ans = 0;    if(l <= mid) ans = query(l, r, k << 1);    if(ans) return ans;    if(r > mid) ans = query(l, r, k << 1|1);    return ans;}int Query(int v, int u){    int t1 = top[v], t2 = top[u];    int ans = -1, tmp = 0;    while(t1 != t2)    {        if(dep[t1] < dep[t2])            swap(t1, t2), swap(v, u);        tmp = query(id[t1], id[v], 1);        if(tmp) ans = tmp;        v = fat[t1], t1 = top[v];    }    if(dep[v] > dep[u]) swap(v, u);    tmp = query(id[v], id[u], 1);    if(tmp) ans = tmp;    return ans;}int main(){    int n, m, x, y;    while(~ scanf("%d%d", &n, &m))    {        init();        for(int i = 1; i <= n-1; i++)        {            scanf("%d%d", &x, &y);            add_edge(x, y);            add_edge(y, x);        }        dfs1(1, 0, 1);        dfs2(1, 1);        build(1, num, 1);        for(int i = 1; i <= n; i++) rid[id[i]] = i;        for(int i = 1; i <= m; i++)        {            scanf("%d%d", &x, &y);            if(x == 0) update(id[y], 1);            else printf("%d\n", Query(x, y));        }    }    return 0;}