HDU 3966 Aragorn's Story (简单树链剖分)

题意：给一棵树，并给定各个点权的值，然后有3种操作：

I C1 C2 K: 把C1与C2的路径上的所有点权值加上K

D C1 C2 K：把C1与C2的路径上的所有点权值减去K

Q C：查询节点编号为C的权值

析：就是简单的树链剖分，可以用树状数组来维护，然后就OK了。

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 = 50000 + 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 edges[maxn<<1];
int p[maxn], fp[maxn], son[maxn], top[maxn];
int dp[maxn], fa[maxn], head[maxn], num[maxn];
int pos, cnt;
int sum[maxn];

void init(){
  pos = 1;  cnt = 0;
  memset(sum, 0, sizeof sum);
  memset(son, -1, sizeof son);
  memset(head, -1, sizeof head);
}
int a[maxn];

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

void dfs1(int u, int f, int d){
  dp[u] = d;  fa[u] = f;
  num[u] = 1;
  for(int i = head[u]; ~i; i = edges[i].next){
    int v = edges[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 = edges[i].next){
    int v = edges[i].to;
    if(v != fa[u] && v != son[u])  dfs2(v, v);
  }
}

int lowbit(int x){ return -x&x; }
void add(int x, int val){
  while(x <= n){
    sum[x] += val;
    x += lowbit(x);
  }
}

int getSum(int x){
  int ans = 0;
  while(x){
    ans += sum[x];
    x -= lowbit(x);
  }
  return ans;
}

void update(int u, int v, int val){
  int f1 = top[u], f2 = top[v];
  while(f1 != f2){
    if(dp[f1] < dp[f2]){
      swap(f1, f2);
      swap(u, v);
    }
    add(p[f1], val);
    add(p[u]+1, -val);
    u = fa[f1];
    f1 = top[u];
  }
  if(dp[u] > dp[v])  swap(u, v);
  add(p[u], val);
  add(p[v]+1, -val);
}

int main(){
  int q;
  while(scanf("%d %d %d", &n, &m, &q) == 3){
    for(int i = 1; i <= n; ++i)  scanf("%d", a+i);
    init();
    for(int i = 0; i < m; ++i){
      int u, v;
      scanf("%d %d", &u, &v);
      add_edge(u, v);
      add_edge(v, u);
    }
    dfs1(1, 0, 0);
    dfs2(1, 1);
    for(int i = 1; i <= n; ++i){
      add(p[i], a[i]);
      add(p[i]+1, -a[i]);
    }
    int u, v, x;
    while(q--){
      char s[5];
      scanf("%s", s);
      if(s[0] == 'Q'){
        scanf("%d", &x);
        printf("%d
", getSum(p[x]));
        continue;
      }
      scanf("%d %d %d", &u, &v, &x);
      if(s[0] == 'D')  x = -x;
      update(u, v, x);
    }
  }
  return 0;
}