hdu 3966 Aragorn's Story(树链剖分+树状数组)

题目链接：hdu 3966 Aragorn's Story

题目大意：给定一个棵树，然后三种操作

Q x：查询节点x的值
I x y w：节点x到y这条路径上全部节点的值添加w
D x y w：节点x到y这条路径上全部节点的值降低w

解题思路：树链剖分，用树状数组维护每一个节点的值。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>

using namespace std;

#define lowbit(x) ((x)&(-x))
const int maxn = 50000;

int N, M, Q, val[maxn+5], fenw[maxn+5];
vector<int> g[maxn+5];

inline void add (int x, int v) {
while (x <= N) {
fenw[x] += v;
x += lowbit(x);
}
}

inline void add (int l, int r, int v) {
add(l, v);
add(r+1, -v);
}

inline int query (int x) {
int ret = 0;
while (x) {
ret += fenw[x];
x -= lowbit(x);
}
return ret;
}

int id, far[maxn+5], son[maxn+5], dep[maxn+5], cnt[maxn+5], top[maxn+5], idx[maxn+5];

void dfs_fir(int u, int pre, int d) {
dep[u] = d;
cnt[u] = 1;
son[u] = 0;
far[u] = pre;

for (int i = 0; i < g[u].size(); i++) {
int v = g[u][i];
if (v == pre)
continue;

dfs_fir(v, u, d + 1);
cnt[u] += cnt[v];
if (cnt[son[u]] < cnt[v])
son[u] = v;
}
}

void dfs_sec(int u, int rot) {
idx[u] = id++;
top[u] = rot;

if (son[u])
dfs_sec(son[u], rot);

for (int i = 0; i < g[u].size(); i++) {
int v = g[u][i];
if (v == son[u] || v == far[u])
continue;
dfs_sec(v, v);
}
}

void change (int u, int v, int c) {
int p = top[u], q = top[v];
while (p != q) {
if (dep[p] < dep[q]) {
swap(p, q);
swap(u, v);
}
add(idx[p], idx[u], c);
u = far[p];
p = top[u];
}

/*
if (dep[u] > dep[v]) swap(u, v);
add(idx[u], idx[v], c);
*/
if (dep[u] < dep[v]) swap(u, v);
add(idx[v], idx[u], c);
}

int main () {
while (scanf("%d%d%d", &N, &M, &Q) == 3) {
id = 1;
memset(fenw, 0, sizeof(fenw));
for (int i = 1; i <= N; i++) {
scanf("%d", &val[i]);
g[i].clear();
}

int u, v, c;
for (int i = 0; i < M; i++) {
scanf("%d%d", &u, &v);
g[u].push_back(v);
g[v].push_back(u);
}

dfs_fir(1, -1, 0);
dfs_sec(1, 1);

for (int i = 1; i <= N; i++) add(idx[i], idx[i], val[i]);

char op[5];
while (Q--) {
scanf("%s%d", op, &u);
if (op[0] == 'Q')
printf("%d\n", query(idx[u]));
else {
scanf("%d%d", &v, &c);
if (op[0] == 'D') c = -c;
change(u, v, c);
}
}
}
return 0;
}