hdu 4897 Little Devil I(树链剖分+线段树)

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题目链接：hdu 4897 Little Devil I

题目大意：给定一棵树，每条边有黑白两种颜色，初始都是白色，现在有三种操作：

1 u v：u到v路径上的边都取成相反的颜色
2 u v：u到v路径上相邻的边都取成相反的颜色（相邻即仅有一个节点在路径上）
3 u v：查询u到v路径上有多少个黑色边

解题思路：树链剖分，用两个线段W和L维护，W对应的是每条的黑白情况，L表示的是每个节点的相邻边翻转情况

（对于轻链而言，重链直接在W上修改）

对于1操作，即为普通的树链剖分，直接在W上修改即可。

对于2操作，每次修改在L上，不过链的两端（即重链的部分）还需要在W上修改

对于3操作，每次查询，重链可以直接根据W中的值判断，轻链还要根据对应节点的L值来确定。

#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 1e5+5;#define lson(x) ((x)<<1)#define rson(x) (((x)<<1)|1)struct Tree {    int lc[maxn << 2], rc[maxn << 2], fp[maxn << 2], s[maxn << 2];    void splay(int u) {        s[u] = rc[u] - lc[u] + 1 - s[u];        fp[u] ^= 1;    }    void pushup(int u) {        s[u] = s[lson(u)] + s[rson(u)];    }    void pushdown(int u) {        if (fp[u]) {            splay(lson(u));            splay(rson(u));            fp[u] = 0;        }    }    void build (int u, int l, int r) {        lc[u] = l;        rc[u] = r;        fp[u] = s[u] = 0;        if (l == r)            return;        int mid = (l + r) / 2;        build(lson(u), l, mid);        build(rson(u), mid + 1, r);        pushup(u);    }    void modify(int u, int l, int r) {        if (l <= lc[u] && rc[u] <= r) {            splay(u);            return;        }        pushdown(u);        int mid = (lc[u] + rc[u]) / 2;        if (l <= mid)            modify(lson(u), l, r);        if (r > mid)            modify(rson(u), l, r);        pushup(u);    }    int query(int u, int l, int r) {        if (l <= lc[u] && rc[u] <= r)            return s[u];        pushdown(u);        int mid = (lc[u] + rc[u]) / 2, ret = 0;        if (l <= mid)            ret += query(lson(u), l, r);        if (r > mid)            ret += query(rson(u), l, r);        pushup(u);        return ret;    }}W, L;int N, Q, E, first[maxn], jump[maxn * 2], link[maxn * 2];int id, idx[maxn], top[maxn], far[maxn], son[maxn], dep[maxn], cnt[maxn];inline void add_Edge(int u, int v) {    link[E] = v;    jump[E] = first[u];    first[u] = E++;}void dfs (int u, int pre, int d) {    far[u] = pre;    son[u] = 0;    cnt[u] = 1;    dep[u] = d;    for (int i = first[u]; i + 1; i = jump[i]) {        int v = link[i];        if (v == pre)            continue;        dfs(v, u, d + 1);        cnt[u] += cnt[v];        if (cnt[son[u]] < cnt[v])            son[u] = v;    }}void dfs (int u, int rot) {    top[u] = rot;    idx[u] = ++id;    if (son[u])        dfs(son[u], rot);    for (int i = first[u]; i + 1; i = jump[i]) {        int v = link[i];        if (v == far[u] || v == son[u])            continue;        dfs(v, v);    }}void init () {    E = id = 0;    memset(first, -1, sizeof(first));    int u, v;    scanf("%d", &N);    for (int i = 1; i < N; i++) {        scanf("%d%d", &u, &v);        add_Edge(u, v);        add_Edge(v, u);    }    dfs(1, 0, 0);    dfs(1, 1);    W.build(1, 1, N);    L.build(1, 1, N);}void modify(int u, int v, int k) {    int p = top[u], q = top[v];    while (p != q) {        if (dep[p] < dep[q]) {            swap(p, q);            swap(u, v);        }        if (k) {            L.modify(1, idx[p], idx[u]);            W.modify(1, idx[p], idx[p]);            W.modify(1, idx[son[u]], idx[son[u]]);        } else            W.modify(1, idx[p], idx[u]);        u = far[p];        p = top[u];    }    if (dep[u] > dep[v])        swap(u, v);    if (k) {        L.modify(1, idx[u], idx[v]);        W.modify(1, idx[u], idx[u]);        W.modify(1, idx[son[v]], idx[son[v]]);    } else {        if (u == v)            return;        W.modify(1, idx[son[u]], idx[v]);    }}int query(int u, int v) {    int p = top[u], q = top[v], ret = 0;    while (p != q) {        if (dep[p] < dep[q]) {            swap(p, q);            swap(u, v);        }        if (u != p)            ret += W.query(1, idx[son[p]], idx[u]);        ret += (W.query(1, idx[p], idx[p]) ^ L.query(1, idx[far[p]], idx[far[p]]));        u = far[p];        p = top[u];    }    if (u == v) return ret;    if (dep[u] > dep[v])        swap(u, v);    ret += W.query(1, idx[son[u]], idx[v]);    return ret;}int main () {    int cas;    scanf("%d", &cas);    while (cas--) {        init();        int k, u, v;        scanf("%d", &Q);        while (Q--) {            scanf("%d%d%d", &k, &u, &v);            if (k == 3)                printf("%d\n", query(u, v));            else                modify(u, v, k - 1);        }    }    return 0;}

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