题目:http://acm.hdu.edu.cn/showproblem.php?pid=3966
题意:给定一棵树,树的节点有点权,有P个询问,第一种是两点路径上(包括这两点)的点全部加上某个值,第二种是两点路径上(包括这两点)的点全部减去某个值,第三点是查询某个点的点权
思路:树链剖分啊加线段树lazy操作。lazy操作写挫了,我也是被自己的弱智惊呆了。。。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
const int N = 50010;
struct edge
{
int to, next;
}g[N*2];
struct node
{
int l, r;
ll sum, mark;
}s[N*4];
int dep[N], top[N], siz[N], id[N], fat[N], son[N], head[N];
int tmp[N], val[N];
int n, m, k, num, cnt;
void add_edge(int v, int u)
{
g[cnt].to = u;
g[cnt].next = head[v];
head[v] = cnt++;
}
void dfs1(int v, int fa, int d)
{
dep[v] = d, siz[v] = 1, fat[v] = fa, son[v] = 0;
for(int i = head[v]; i != -1; i = g[i].next)
{
int u = g[i].to;
if(u != fa)
{
dfs1(u, v, d + 1);
siz[v] += siz[u];
if(siz[son[v]] < siz[u]) son[v] = u;
}
}
}
void dfs2(int v, int tp)
{
top[v] = tp, id[v] = ++num;
if(son[v]) dfs2(son[v], tp);
for(int i = head[v]; i != -1; i = g[i].next)
{
int u = g[i].to;
if(u != fat[v] && u != son[v]) dfs2(u, u);
}
}
void push_down(int k)
{
if(s[k].mark)
{
s[k<<1].mark += s[k].mark;
s[k<<1|1].mark += s[k].mark;
s[k<<1].sum += s[k].mark;
s[k<<1|1].sum += s[k].mark;
s[k].mark = 0;
}
}
void build(int l, int r, int k)
{
s[k].l = l, s[k].r = r, s[k].mark = 0;
if(l == r)
{
s[k].sum = val[l];
return;
}
int mid = (l + r) >> 1;
build(l, mid, k << 1);
build(mid + 1, r, k << 1|1);
}
void update(int l, int r, int c, int k)
{
if(l <= s[k].l && s[k].r <= r)
{
s[k].sum += c;
s[k].mark += c;
return;
}
push_down(k);
int mid = (s[k].l + s[k].r) >> 1;
if(l <= mid) update(l, r, c, k << 1);
if(r > mid) update(l, r, c, k << 1|1);
}
void renew(int v, int u, int c)
{
int t1 = top[v], t2 = top[u];
while(t1 != t2)
{
if(dep[t1] < dep[t2])
swap(t1, t2), swap(v, u);
update(id[t1], id[v], c, 1);
v = fat[t1], t1 = top[v];
}
if(dep[v] > dep[u]) swap(v, u);
update(id[v], id[u], c, 1);
}
ll query(int x, int k)
{
if(s[k].l == s[k].r)
return s[k].sum;
push_down(k);
int mid = (s[k].l + s[k].r) >> 1;
if(x <= mid) return query(x, k << 1);
else return query(x, k << 1|1);
}
void slove()
{
char ch;
int a, b, c;
while(k--)
{
scanf(" %c", &ch);
if(ch == 'I')
{
scanf("%d%d%d", &a, &b, &c);
renew(a, b, c);
}
else if(ch == 'D')
{
scanf("%d%d%d", &a, &b, &c);
renew(a, b, -c);
}
else if(ch == 'Q')
{
scanf("%d", &a);
printf("%I64d\n", query(id[a], 1));
}
}
}
int main()
{
int a, b;
while(~ scanf("%d%d%d", &n, &m, &k))
{
num = cnt = 0;
memset(head, -1, sizeof head);
for(int i = 1; i <= n; i++)
scanf("%d", tmp + i);
for(int i = 1; i <= m; i++)
{
scanf("%d%d", &a, &b);
add_edge(a, b);
add_edge(b, a);
}
dfs1(1, 0, 1);
dfs2(1, 1);
for(int i = 1; i <= n; i++)
val[id[i]] = tmp[i];
build(1, num, 1);
slove();
}
return 0;
}