HDU 3966 Aragorn's Story(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。

思路：树链剖分.............一开始线段树写sb了wa了一下午.............正好又练习了一下线段树........

#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<ctime>
#define eps 1e-6
#define LL long long
#define pii (pair<int, int>)
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const int maxn = 50005;
const int INF = 1400000000;
int addv[maxn*4], maxv[maxn*4];
int W[maxn]; 
void maintain(int o, int L, int R) {
	int lc = o*2, rc = o*2+1;
	maxv[o] = 0;
	if(R > L) {   //考虑左右子树 
		maxv[o] = max(maxv[lc], maxv[rc]);
	}
	maxv[o] += addv[o];//考虑add操作 
} 
void update(int o, int L, int R, int v, int yl, int yr) {
	int lc = o*2, rc = o*2+1;
	if(yl <= L && yr >= R) {   //递归边界 
		addv[o] += v;		//累加边界的add值 
	} else {
		int M = L + (R-L)/2;
		if(yl <= M) update(lc, L, M, v, yl, yr);
		if(yr > M) update(rc, M+1, R, v, yl, yr);
	}
	maintain(o, L, R);		//递归结束前重新计算本节点的附加信息 
} 
int query(int o, int L, int R, int add, int yl, int yr) {
	int ans = -INF;
	if(yl <= L && yr >= R) {
		return maxv[o] + add;
	} else {
		int M = L + (R-L)/2;
		if(yl <= M) ans = max(ans, query(o*2, L, M, add + addv[o], yl, yr));
		if(yr > M) ans = max(ans, query(o*2+1, M+1, R, add + addv[o], yl, yr));
	}
	return ans;
} 
int n, gan, q, tot;
vector<int> G[maxn];
int siz[maxn], son[maxn], dep[maxn], top[maxn], fa[maxn], pos[maxn], arc[maxn];
void init() {
	for(int i = 1; i <= n; i++) G[i].clear();
	tot = 0;
	memset(son, 0, sizeof(son));
	memset(addv, 0, sizeof(addv));
	memset(maxv, 0, sizeof(maxv));
}
void dfs(int cur, int f) {
	siz[cur] = 1;
	int tmp = 0;
	for(int i = 0; i < G[cur].size(); i++) {
		int u = G[cur][i];
		if(u == f) continue;
		dep[u] = dep[cur] + 1;
		fa[u] = cur;
		dfs(u, cur);
		siz[cur] += siz[u];
		if(siz[u] > tmp) son[cur] = u, tmp = siz[u];
	}
}
void dfs2(int cur, int tp) {
	top[cur] = tp;
	pos[cur] = ++tot;
	arc[tot] = cur;
	if(son[cur]) dfs2(son[cur], tp);
	for(int i = 0; i < G[cur].size(); i++) {
		int u = G[cur][i];
		if(u==son[cur] || u==fa[cur]) continue;
		dfs2(u, u);
	}
}
void modify(int u, int v, int d) {
	int fu = top[u], fv = top[v];
	while(fu != fv) {
		if(dep[fu]<dep[fv]) swap(fu, fv), swap(u, v);
		update(1, 1, n, d, pos[fu], pos[u]);
		u = fa[fu]; fu = top[u];
	}
	if(dep[u]<dep[v]) swap(u, v);
	update(1, 1, n, d, pos[v], pos[u]);
}
void Build(int o, int L, int R) {
	if(L == R) maxv[o]=W[arc[L]], addv[o]+=maxv[o];
	else {
		int M = L + (R-L)/2;
		Build(2*o, L, M); Build(2*o+1, M+1, R);
		maxv[o] = max(maxv[2*o], maxv[2*o+1]);
	}
}
int main() {
    //freopen("input.txt", "r", stdin);
	while(scanf("%d%d%d", &n, &gan, &q)==3) {
		init();
		for(int i = 1; i <= n; i++) scanf("%d", &W[i]);
		for(int i = 1; i <= n-1; i++) {
			int u, v; scanf("%d%d", &u, &v);
			G[u].push_back(v);
			G[v].push_back(u);
		}
		dfs(1, 0);
		dfs2(1, 1);
		Build(1, 1, n);		
		char op[10];
		while(q--) {
			scanf("%s", op);
			int u, v, d;
			if(op[0] == 'Q') {
				scanf("%d", &u);
				printf("%d\n", query(1, 1, n, 0, pos[u], pos[u]));
			}
			else if(op[0] == 'D'){
				scanf("%d%d%d", &u, &v, &d);
				modify(u, v, -d);
			}
			else {
				scanf("%d%d%d", &u, &v, &d);
				modify(u, v, d);
			}
		}
	}
    return 0;
}