HDU - 3966 Aragorn's Story（树链剖分模板题）

Our protagonist is the handsome human prince Aragorn comes from The Lord of the Rings. One day Aragorn finds a lot of enemies who want to invade his kingdom. As Aragorn knows, the enemy has N camps out of his kingdom and M edges connect them. It is guaranteed that for any two camps, there is one and only one path connect them. At first Aragorn know the number of enemies in every camp. But the enemy is cunning , they will increase or decrease the number of soldiers in camps. Every time the enemy change the number of soldiers, they will set two camps C1 and C2. Then, for C1, C2 and all camps on the path from C1 to C2, they will increase or decrease K soldiers to these camps. Now Aragorn wants to know the number of soldiers in some particular camps real-time.

Input
Multiple test cases, process to the end of input.

For each case, The first line contains three integers N, M, P which means there will be N(1 ≤ N ≤ 50000) camps, M(M = N-1) edges and P(1 ≤ P ≤ 100000) operations. The number of camps starts from 1.

The next line contains N integers A1, A2, …AN(0 ≤ Ai ≤ 1000), means at first in camp-i has Ai enemies.

The next M lines contains two integers u and v for each, denotes that there is an edge connects camp-u and camp-v.

The next P lines will start with a capital letter ‘I’, ‘D’ or ‘Q’ for each line.

‘I’, followed by three integers C1, C2 and K( 0≤K≤1000), which means for camp C1, C2 and all camps on the path from C1 to C2, increase K soldiers to these camps.

‘D’, followed by three integers C1, C2 and K( 0≤K≤1000), which means for camp C1, C2 and all camps on the path from C1 to C2, decrease K soldiers to these camps.

‘Q’, followed by one integer C, which is a query and means Aragorn wants to know the number of enemies in camp C at that time.

Output
For each query, you need to output the actually number of enemies in the specified camp.

存模板，只询问单点，不需要pushUp，pushDown也只用来节省时间

#include<bits/stdc++.h>
using namespace std;

const int maxn = 50005;
vector<int> G[maxn];
int n, m, q;
int in[maxn];
//子节点个数，深度
int sz[maxn], dep[maxn];
//重儿子，父节点
int ch[maxn], fa[maxn];
//重链开头,dfs序，原序
int top[maxn], tid[maxn], tid2[maxn];
int sum[maxn << 2], lazy[maxn << 2];
int tot;


void dfs1(int u, int f, int d) {
	sz[u] = 1;
	fa[u] = f;
	dep[u] = d;
	for(int i = 0; i < G[u].size(); i++) {
		int v = G[u][i];
		if(v == f) {
			continue;
		}
		dfs1(v, u, d + 1);
		sz[u] += sz[v];
		if(ch[u] == -1 || sz[v] > sz[ch[u]]) {
			ch[u] = v;
		}
	}
}

void dfs2(int u, int tp) {
	top[u] = tp;
	tid[u] = ++tot;
	tid2[tot] = u;
	if(ch[u] == -1) {
		return;
	}
	dfs2(ch[u], tp);
	for(int i = 0; i < G[u].size(); i++) {
		int v = G[u][i];
		if(v != ch[u] && v != fa[u]) {
			dfs2(v, v);
		}
	}
}

void pushDown(int i, int len) {
	if(!lazy[i]) {
		return;
	}
	lazy[i << 1] += lazy[i];
	lazy[i << 1 | 1] += lazy[i];
	sum[i << 1] += (len - (len >> 1)) * lazy[i];
	sum[i << 1 | 1] += (len >> 1) * lazy[i];
	lazy[i] = 0;
}

void build(int i, int l, int r) {
	lazy[i] = 0;
	if(l == r) {
		// 以重链优先，dfs顺序加入线段树
		sum[i] = in[tid2[l]];
		return;
	}
	int mid = (l + r) >> 1;
	build(i << 1, l, mid);
	build(i << 1 | 1, mid + 1, r);
}

void update(int i, int l, int r, int L, int R, int val) {
	if(l >= L && r <= R) {
		sum[i] += (r - l + 1) * val;
		lazy[i] += val;
		return;
	}
	pushDown(i, r - l + 1);
	int mid = (l + r) >> 1;
	if(L <= mid) {
		update(i << 1, l, mid, L, R, val);
	}
	if(R > mid) {
		update(i << 1 | 1, mid + 1, r, L, R, val);
	}
}

int query(int i, int l, int r, int pos) {
	if(l == r) {
		return sum[i];
	}
	pushDown(i, r - l + 1);
	int mid = (l + r) >> 1;
	int ans;
	if(pos <= mid) {
		ans = query(i << 1, l, mid, pos);
	} else {
		ans = query(i << 1 | 1, mid + 1, r, pos);
	}
	return ans;
}

void solve(int u, int v, int val) {
	int f1 = top[u], f2 = top[v];
	while(f1 != f2) {
		if(dep[f1] < dep[f2]) {
			swap(f1, f2);
			swap(u, v);
		}
		update(1, 1, n, tid[f1], tid[u], val);
		u = fa[f1];
		f1 = top[u];
	}
	if(dep[u] < dep[v]) {
		swap(u, v);
	}
	update(1, 1, n, tid[v], tid[u], val);
}

/*
9 8 99
1 1 1 1 1 1 1 1 1
1 2
2 3
1 4
4 5
4 6
6 8
6 7
8 9
*/


int main() {
	while(~scanf("%d%d%d", &n, &m, &q)) {
		memset(ch, -1, sizeof(ch));
		for(int i = 1; i <= n; i++) {
			G[i].clear();
		}
		for(int i = 1; i <= n; i++) {
			scanf("%d", &in[i]);
		}

		int u, v, val;
		while(m--) {
			scanf("%d%d", &u, &v);
			G[u].push_back(v);
			G[v].push_back(u);
		}

		tot = 0;
		dfs1(1, 0, 0);
		dfs2(1, 1);
		build(1, 1, n);

		char op[3];
		while(q--) {
			scanf("%s", op);
			if(op[0] == 'Q') {
				scanf("%d", &u);
				printf("%d\n", query(1, 1, n, tid[u]));
			} else {
				scanf("%d%d%d", &u, &v, &val);
				if(op[0] == 'D') {
					val = -val;
				}
				solve(u, v, val);
			}
		}
	}
	return 0;
}