POJ 3237 Tree（树链剖分）

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题目链接：点击打开链接
思路：
对于树上的路径更新操作，我们通常把他hash到线段上，也就是树链剖分，大概完全理解了吧，存个代码。
对于该题的反转操作，可以里用异或操作的性质来做标记。
细节参见代码：
#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <string>#include <vector>#include <stack>#include <ctime>#include <bitset>#include <cstdlib>#include <cmath>#include <set>#include <list>#include <deque>#include <map>#include <queue>#define Max(a,b) ((a)>(b)?(a):(b))#define Min(a,b) ((a)<(b)?(a):(b))using namespace std;typedef long long ll;typedef long double ld;const double eps = 1e-6;const double PI = acos(-1);const int mod = 1000000000 + 7;const int INF = 0x3f3f3f3f;const int seed = 131;const ll INF64 = ll(1e18);const int maxn = 10000 + 10;int T,n,m,tree_id = 0,minv[maxn<<2],maxv[maxn<<2], setv[maxn<<2],val[maxn];struct node {    int u, d;    node(int u=0, int d=0):u(u),d(d) {}};vector<node> g[maxn];int pre[maxn]; /// x的父亲int siz[maxn]; /// x的子树规模int son[maxn]; /// x的重儿子int dep[maxn]; /// x相对于根结点的深度void dfs(int u, int fa) {    siz[u] = 1;    pre[u] = fa ;    dep[u] = dep[fa] + 1;    son[u] = 0;    int len = g[u].size(), maxv = 0;    for(int i = 0; i < len; i++) {        int v = g[u][i].u;        if(v == fa) continue;        dfs(v, u);        siz[u] += siz[v];        if(siz[v] > maxv) {            maxv = siz[v];            son[u] = v;        }    }}int top[maxn]; /// 这条重链的头部int pos[maxn]; /// x重标号后的标号/// tree_idx 用以给所有边重标号void build_tree(int u, int top_id) {    top[u] = top_id;    pos[u] = ++tree_id;    if(son[u]) build_tree(son[u], top_id);    int len = g[u].size();    for(int i = 0; i < len; i++) {        int v = g[u][i].u;        if(v == pre[u]) continue;        if(v != son[u]) {            build_tree(v, v);        }    }}void pushup(int o) {    minv[o] = min(minv[o<<1], minv[o<<1|1]);    maxv[o] = max(maxv[o<<1], maxv[o<<1|1]);}void pushdown(int l, int r, int o) {    if(l == r) return ;    if(setv[o]) {        setv[o] ^= 1;        setv[o<<1] ^= 1;        int t = minv[o<<1];        minv[o<<1] = -maxv[o<<1];        maxv[o<<1] = -t;        setv[o<<1|1] ^= 1;        t = minv[o<<1|1];        minv[o<<1|1] = -maxv[o<<1|1];        maxv[o<<1|1] = -t;    }}void build(int l, int r, int o) {    minv[o] = INF;    maxv[o] = -INF;    setv[o] = 0;    if(l == r) {        minv[o] = maxv[o] = val[l];        return ;    }    int mid = (l + r) >> 1;    build(l, mid, o<<1);    build(mid+1, r, o<<1|1);    pushup(o);}void update(int L, int R, int v, int l, int r, int o) {    if(L <= l && r <= R) {        if(v == INF) {            setv[o] ^= 1;            int t = minv[o]; minv[o] = -maxv[o]; maxv[o] = -t;        }        else {            minv[o] = maxv[o] = v; setv[o] = 0;        }        return ;    }    pushdown(l, r, o);    int mid = (l + r) >> 1;    if(L <= mid) update(L, R, v, l, mid, o<<1);    if(mid < R) update(L, R, v, mid+1, r, o<<1|1);    pushup(o);}int query(int L, int R, int l, int r, int o) {    if(L <= l && r <= R) {        return maxv[o];    }    int mid = (l + r) >> 1;    int ans = -INF;    pushdown(l, r, o);    if(L <= mid) ans = max(ans, query(L, R, l, mid, o<<1));    if(mid < R) ans = max(ans, query(L, R, mid+1, r, o<<1|1));    pushup(o);    return ans;}int solve(int x, int y) {    int res = -INF;    while(top[x] != top[y]) {        if(dep[top[x]] < dep[top[y]]) swap(x, y);        res = max(res, query(pos[top[x]], pos[x], 1, n, 1));        x = pre[top[x]];    }    if(x == y) return res;    if(dep[x] > dep[y]) swap(x, y);    return max(res, query(pos[x]+1, pos[y], 1, n, 1));}void NEGATE(int x, int y) {    while(top[x] != top[y]) {        if(dep[top[x]] < dep[top[y]]) swap(x, y);        update(pos[top[x]], pos[x], INF, 1, n, 1);        x = pre[top[x]];    }    if(x == y) return ;    if(dep[x] > dep[y]) swap(x, y);    update(pos[x]+1, pos[y], INF, 1, n, 1);}void init() {    tree_id = 0;    for(int i = 1; i <= n; i++) g[i].clear();}struct edge {    int a, b, c;}e[maxn];int main() {    int T; scanf("%d", &T);    while(T--) {        scanf("%d", &n);        init();        for(int i = 1; i < n; i++) {            scanf("%d%d%d", &e[i].a, &e[i].b, &e[i].c);            g[e[i].a].push_back(node(e[i].b, e[i].c));            g[e[i].b].push_back(node(e[i].a, e[i].c));        }        dfs(1, 0);        build_tree(1, 1);        for(int i = 1; i < n; i++) {            int a = e[i].a, b = e[i].b;            if(dep[a] < dep[b]) swap(a, b);            val[pos[a]] = e[i].c;        }        build(1, n, 1);        while(true) {            char op[10]; scanf("%s", op);            int a, b;            if(op[0] != 'D') scanf("%d%d", &a, &b);            if(op[0] == 'D') break;            else if(op[0] == 'Q') {                printf("%d\n", solve(a, b));            }            else if(op[0] == 'C') {                int x = e[a].a, y = e[a].b;                if(dep[x] < dep[y]) swap(x, y);                update(pos[x], pos[x], b, 1, n, 1);            }            else {                NEGATE(a, b);            }        }    }    return 0;}
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