HYSBZ 1036 树的统计Count (水题树链剖分)

题意：中文题。

析：就是直接维护一个最大值和一个和，用线段树维护即可，这个题很简单，但是我卡了一晚上，就是在定位的时候，位置直接反过来了，但是样例全过了。。。真是。。。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 30000 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}

struct Edge{
  int to, next;
};
Edge edge[maxn<<1];
int head[maxn], cnt, fa[maxn];
int dep[maxn], top[maxn], num[maxn];
int son[maxn], p[maxn], pos, a[maxn], fp[maxn];
int sum[maxn<<2], maxv[maxn<<2];

void init(){
  memset(son, -1, sizeof son);
  cnt = 0; pos = 1;
  memset(head, -1, sizeof head);
}

void add_edge(int u, int v){
  edge[cnt].to = v;
  edge[cnt].next = head[u];
  head[u] = cnt++;
}

void dfs1(int u, int f, int d){
  dep[u] = d;  fa[u] = f;
  num[u] = 1;
  for(int i = head[u]; ~i; i = edge[i].next){
    int v = edge[i].to;
    if(v == f)  continue;
    dfs1(v, u, d+1);
    num[u] += num[v];
    if(son[u] == -1 || num[v] > num[son[u]])  son[u] = v;
  }
}

void dfs2(int u, int sp){
  top[u] = sp; p[u] = pos++;
  fp[p[u]] = u;
  if(son[u] == -1)  return ;
  dfs2(son[u], sp);
  for(int i = head[u]; ~i; i = edge[i].next){
    int v = edge[i].to;
    if(v == fa[u] || v == son[u])  continue;
    dfs2(v, v);
  }
}

void push_up(int rt){
  sum[rt] = sum[rt<<1] + sum[rt<<1|1];
  maxv[rt] = max(maxv[rt<<1], maxv[rt<<1|1]);
}

void build(int l, int r, int rt){
  if(l == r){
    sum[rt] = maxv[rt] = a[fp[l]];
    return ;
  }
  int m = l + r >> 1;
  build(lson);
  build(rson);
  push_up(rt);
}

void update(int M, int val, int l, int r, int rt){
  if(l == r){
    sum[rt] = maxv[rt] = val;
    return ;
  }
  int m = l + r >> 1;
  if(M <= m)  update(M, val, lson);
  else  update(M, val, rson);
  push_up(rt);
}

int queryMax(int L, int R, int l, int r, int rt){
  if(L <= l && r <= R)  return maxv[rt];
  int m = l + r >> 1;
  int ans = -INF;
  if(L <= m)  ans = queryMax(L, R, lson);
  if(R > m)   ans = max(ans, queryMax(L, R, rson));
  return ans;
}

int querySum(int L, int R, int l, int r, int rt){
  if(L <= l && r <= R)  return sum[rt];
  int m = l + r >> 1;
  int ans = 0;
  if(L <= m)  ans = querySum(L, R, lson);
  if(R > m)   ans += querySum(L, R, rson);;
  return ans;
}

int solve(int u, int v, bool ok){
  int f1 = top[u], f2 = top[v];
  int ans = -INF;
  if(ok) ans = 0;
  while(f1 != f2){
    if(dep[f1] < dep[f2]){
      swap(f1, f2);
      swap(u, v);
    }
    if(ok)  ans += querySum(p[f1], p[u], 1, n, 1);
    else  ans = max(ans, queryMax(p[f1], p[u], 1, n, 1));
    u = fa[f1];
    f1 = top[u];
  }
  if(dep[u] > dep[v])  swap(u, v);
  if(ok)  return ans + querySum(p[u], p[v], 1, n, 1);
  return max(ans, queryMax(p[u], p[v], 1, n, 1));
}

int main(){
  while(scanf("%d", &n) == 1){
    init();
    for(int i = 1; i < n; ++i){
      int u, v;
      scanf("%d %d", &u, &v);
      add_edge(u, v);
      add_edge(v, u);
    }
    for(int i = 1; i <= n; ++i)  scanf("%d", a+i);
    dfs1(1, 0, 0);
    dfs2(1, 1);
    build(1, n, 1);
    scanf("%d", &m);
    while(m--){
      char op[10];
      int u, v;
      scanf("%s %d %d", op, &u, &v);
      if(op[0] == 'C')  update(p[u], v, 1, n, 1);
      else  printf("%d\n", solve(u, v, op[1] == 'S'));
    }
  }
  return 0;
}