bzoj 1036: [ZJOI2008]树的统计Count（树链剖分点权）

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

solution：

树链剖分点权模板题。

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define inf -1e9
#define ls (t<<1)
#define rs (t<<1)|1
const int maxn = 30010;
int n,no, dfsxu, val[maxn],pos[maxn],head[maxn],heavy[maxn],top[maxn],idx[maxn],dep[maxn],fa[maxn],size[maxn];
struct Edge
{int to, next;
}edge[maxn<<1];
struct segtree
{int l, r, sum,max;
}tree[maxn<<2];
void init()
{no = dfsxu = 0;memset(head, -1, sizeof(head));memset(heavy, 0, sizeof(heavy));dep[1] = fa[1] =size[0]=0;
}
void add(int u, int v)
{edge[no].to = v;edge[no].next = head[u];head[u] = no++;edge[no].to = u;edge[no].next = head[v];head[v] = no++;
}
void dfs(int x)
{size[x] = 1;for (int i = head[x]; ~i; i = edge[i].next){int u = edge[i].to;if (u != fa[x]){fa[u] = x;dep[u] = dep[x] + 1;dfs(u);size[x] += size[u];if (size[heavy[x]] < size[u])heavy[x] = u;}}
}
void spilt(int x, int y)//剖分
{top[x] = y;idx[x] = ++dfsxu;pos[dfsxu] = x;if (heavy[x])spilt(heavy[x], top[x]);for (int i = head[x]; ~i; i = edge[i].next){int u = edge[i].to;if (u != fa[x] && u != heavy[x])spilt(u, u);}
}
void pushup(int t)
{tree[t].max = max(tree[ls].max, tree[rs].max);tree[t].sum = tree[ls].sum + tree[rs].sum;
}
void build(int l, int r, int t)
{tree[t].l = l; tree[t].r = r; tree[t].max = inf; tree[t].sum=0;if (l == r){tree[t].max = tree[t].sum = val[pos[l]];return;}int mid = (l + r) >> 1;build(l, mid,ls);build(mid+1, r, rs);pushup(t);
}
void update(int x, int val, int t)
{if (tree[t].l == x&&tree[t].r == x){tree[t].max = tree[t].sum = val;return;}if (x <= tree[ls].r)update(x, val, ls);else update(x, val, rs);pushup(t);
}
int qqsum(int l, int r, int t)
{if (tree[t].l == l&&tree[t].r == r)return tree[t].sum;if (r <= tree[ls].r)return qqsum(l, r, ls);else if (l >= tree[rs].l)return qqsum(l, r, rs);return qqsum(l, tree[ls].r, ls)+qqsum(tree[rs].l, r, rs);
}
int qsum(int x, int y)
{int ans = 0,u = top[x], v = top[y];while (u != v){if (dep[u] < dep[v]){swap(u, v);swap(x, y);}ans += qqsum(idx[u], idx[x], 1);x = fa[u];u = top[x];}if (dep[x] > dep[y])swap(x, y);ans += qqsum(idx[x], idx[y], 1);return ans;
}
int qqmax(int l, int r, int t)
{if (tree[t].l == l&& tree[t].r == r)return tree[t].max;if (r <= tree[ls].r)return qqmax(l, r, ls);else if (l >= tree[rs].l)return qqmax(l, r, rs);return max(qqmax(l, tree[ls].r, ls), qqmax(tree[rs].l, r, rs));
}
int qmax(int x, int y)
{int ans = inf, u = top[x], v = top[y];while (u != v){if (dep[u] < dep[v]){swap(u, v);swap(x, y);}ans = max(ans,qqmax(idx[u], idx[x], 1));x = fa[u];u = top[x];}if (dep[x] > dep[y])swap(x, y);ans = max(ans,qqmax(idx[x], idx[y], 1));return ans;
}
int main()
{int x,y,q;char op[10];scanf("%d", &n);init();for (int i = 1; i < n; i++){scanf("%d%d", &x, &y);add(x, y);}for (int i = 1; i <= n; i++)scanf("%d", &val[i]);dfs(1);spilt(1, 1);build(1, n, 1);scanf("%d", &q);while (q--){scanf("%s%d%d", op,&x,&y);if (op[0] == 'C')update(idx[x], y, 1);else if (op[1] == 'S')printf("%d\n", qsum(x,y));else printf("%d\n", qmax(x, y));}return 0;
}