BZOJ 1036 树的统计

Description

　　一棵树上有n个节点，编号分别为1到n，每个节点都有一个权值w。我们将以下面的形式来要求你对这棵树完成
一些操作： I. CHANGE u t : 把结点u的权值改为t II. QMAX u v: 询问从点u到点v的路径上的节点的最大权值 I
II. QSUM u v: 询问从点u到点v的路径上的节点的权值和 注意：从点u到点v的路径上的节点包括u和v本身

Input

　　输入的第一行为一个整数n，表示节点的个数。接下来n – 1行，每行2个整数a和b，表示节点a和节点b之间有
一条边相连。接下来n行，每行一个整数，第i行的整数wi表示节点i的权值。接下来1行，为一个整数q，表示操作
的总数。接下来q行，每行一个操作，以“CHANGE u t”或者“QMAX u v”或者“QSUM u v”的形式给出。 
对于100％的数据，保证1<=n<=30000，0<=q<=200000；中途操作中保证每个节点的权值w在-30000到30000之间。

Output

　　对于每个“QMAX”或者“QSUM”的操作，每行输出一个整数表示要求输出的结果。

Sample Input

4
1 2
2 3
4 1
4 2 1 3
12
QMAX 3 4
QMAX 3 3
QMAX 3 2
QMAX 2 3
QSUM 3 4
QSUM 2 1
CHANGE 1 5
QMAX 3 4
CHANGE 3 6
QMAX 3 4
QMAX 2 4
QSUM 3 4

Sample Output

4
1
2
2
10
6
5
6
5
16

CNM这不裸的树剖吗..

这几天树剖都快尼玛做吐了...

#include <cstdio>#include <iostream>#include <cstring>#define inf 0x3f3f3f3f#define travel(x) for(int i = head[x]; i; i = edge[i].next)using namespace std;const int MAXN = 30000 + 10;inline int read() {    int x = 0, f = 1; char ch = getchar();    while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}    while(ch >= '0' && ch <= '9') {x = x * 10 + ch - '0'; ch = getchar();}    return x * f;}struct data { int to, next; }edge[MAXN << 1];int head[MAXN], dep[MAXN], size[MAXN], fa[MAXN];int pos[MAXN], top[MAXN], cnt, sz;inline void addedge(int u, int v) {    edge[++cnt].to = v; edge[cnt].next = head[u];    head[u] = cnt;}inline void dfs1(int x) {    size[x] = 1;    travel(x) {    int v = edge[i].to;        if(v == fa[x]) continue;        dep[v] = dep[x] + 1;        fa[v] = x; dfs1(v);        size[x] += size[v];    }}inline void dfs2(int x, int chain) {    int k = 0; sz++; pos[x] = sz; top[x] = chain;    travel(x) { int v = edge[i].to; if(dep[v] > dep[x] && size[v] > size[k]) k = v; }    if(!k) return;    dfs2(k, chain);    travel(x) { int v = edge[i].to; if(dep[v] > dep[x] && k != v) dfs2(v, v); }}struct node { int l, r, mx, sum; }tree[MAXN << 2];int v[MAXN], n, q;inline void pushup(int id) {    tree[id].sum = tree[id << 1].sum + tree[id << 1 | 1].sum;    tree[id].mx = max(tree[id << 1].mx, tree[id << 1 | 1].mx);}inline void build_tree(int id, int l, int r) {    tree[id].l = l; tree[id].r = r; tree[id].mx = tree[id].sum = 0;    if(l == r) return;    int mid = (l + r) >> 1;    build_tree(id << 1, l, mid);    build_tree(id << 1 | 1, mid + 1, r);}inline void update(int id, int pos, int c) {    int l = tree[id].l, r = tree[id].r;    if(l == r) {        tree[id].sum = tree[id].mx = c;        return;    }    int mid = (l + r) >> 1;    if(pos <= mid) update(id << 1, pos, c);    else update(id << 1 | 1, pos, c);    pushup(id);}inline int query_sum(int id, int L, int R) {    int l = tree[id].l, r = tree[id].r;    if(l == L && r == R) return tree[id].sum;    int mid = (l + r) >> 1;    if(R <= mid) return query_sum(id << 1, L, R);    else if(L > mid) return query_sum(id << 1 | 1, L, R);    else return query_sum(id << 1, L, mid) + query_sum(id << 1 | 1, mid + 1, R);}inline int query_max(int id, int L, int R) {    int l = tree[id].l, r = tree[id].r;    if(l == L && r == R) return tree[id].mx;    int mid = (l + r) >> 1;    if(R <= mid) return query_max(id << 1, L, R);    else if(L > mid) return query_max(id << 1 | 1, L, R);    else return max(query_max(id << 1, L, mid), query_max(id << 1 | 1, mid + 1, R));}inline int solvesum(int x, int y) {    int ret = 0;    while(top[x] != top[y]) {        if(dep[top[x]] < dep[top[y]]) swap(x, y);        ret += query_sum(1, pos[top[x]], pos[x]);        x = fa[top[x]];    }    if(pos[x] > pos[y]) swap(x, y);    ret += query_sum(1, pos[x], pos[y]);    return ret;}inline int solvemax(int x, int y) {    int mx = -inf;    while(top[x] != top[y]) {        if(dep[top[x]] < dep[top[y]]) swap(x, y);        mx = max(mx, query_max(1, pos[top[x]], pos[x]));        x = fa[top[x]];    }    if(pos[x] > pos[y]) swap(x, y);    mx = max(mx, query_max(1, pos[x], pos[y]));    return mx;}int main() {    n = read(); q = read();    for(int i = 1; i <= n; ++i) v[i] = read();    for(int i = 1, x, y; i < n; ++i) {        x = read(), y = read();        addedge(x, y);        addedge(y, x);    }    dfs1(1); dfs2(1, 1); build_tree(1, 1, n);    for(int i = 1; i <= n; ++i) update(1, pos[i], v[i]);    char str[10];    for(int i = 1, x, y; i <= q; ++i) {        scanf("%s", str);        if(str[0] == 'C') {            x = read(), y = read();            update(1, pos[x], y);        }        else {            x = read(), y = read();            if(str[1] == 'M') printf("%d\n", solvemax(x, y));            else printf("%d\n", solvesum(x, y));        }    }    return 0;}