POJ 3237 Tree 树链剖分

题目：http://poj.org/problem?id=3237

题意：给定一棵树，有边权，然后有一定数量的操作：第一种操作是改变第i条边（输入顺序）的边权为v，第二种操作是对某两点路径上的所有边权取反，第三种操作求两点间路径的最大权值

思路：树链剖分啊，然后线段树维护区间最大值和最小值，因为取反操作会使最大变最小，最小变最大，在更新最大最小值注意，一不小心就哇了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

typedef long long ll;
const int INF = 0x3f3f3f3f;
const int N = 10010;
struct edge
{
int to, next;
}g[N*2];
struct node
{
int l, r, maxx, minn, mark;
}s[N*4];
int dep
, top
, siz
, id
, fat
, son
, head
;
int d
[3], val
;
int n, 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_up(int k)
{
s[k].maxx = max(s[k<<1].maxx, s[k<<1|1].maxx);
s[k].minn = min(s[k<<1].minn, s[k<<1|1].minn);
}
void push_down(int k)
{
int tmp;
if(s[k].mark & 1)
{
s[k<<1].mark += s[k].mark;
s[k<<1|1].mark += s[k].mark;
/*最大变最小，最小变最大*/
tmp = -s[k<<1].maxx;
s[k<<1].maxx = -s[k<<1].minn;
s[k<<1].minn = tmp;
tmp = -s[k<<1|1].maxx;
s[k<<1|1].maxx = -s[k<<1|1].minn;
s[k<<1|1].minn = tmp;
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].maxx = s[k].minn = val[l]; return;
}
int mid = (l + r) >> 1;
build(l, mid, k << 1);
build(mid + 1, r, k << 1|1);
push_up(k);
}
void update(int l, int r, int k)
{
if(l <= s[k].l && s[k].r <= r)
{
/*最大变最小，最小变最大*/
int tmp = -s[k].maxx;
s[k].maxx = -s[k].minn;
s[k].minn = tmp;
s[k].mark++;
return;
}
push_down(k);
int mid = (s[k].l + s[k].r) >> 1;
if(l <= mid) update(l, r, k << 1);
if(r > mid) update(l, r, k << 1|1);
push_up(k);
}
void renew(int v, int u)
{
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], 1);
v = fat[t1], t1 = top[v];
}
if(v == u) return;
if(dep[v] > dep[u]) swap(v, u);
update(id[son[v]], id[u], 1);
}
void update_edge(int x, int c, int k)
{
if(s[k].l == s[k].r)
{
s[k].maxx = s[k].minn = c;
return;
}
push_down(k);
int mid = (s[k].l + s[k].r) >> 1;
if(x <= mid) update_edge(x, c, k << 1);
else update_edge(x, c, k << 1|1);
push_up(k);
}
int query(int l, int r, int k)
{
if(l <= s[k].l && s[k].r <= r)
return s[k].maxx;
push_down(k);
int mid = (s[k].l + s[k].r) >> 1;
int ans = -INF;
if(l <= mid) ans = max(ans, query(l, r, k << 1));
if(r > mid) ans = max(ans, query(l, r, k << 1|1));
push_up(k);
return ans;
}
int seek(int v, int u)
{
int t1 = top[v], t2 = top[u], ans = -INF;
while(t1 != t2)
{
if(dep[t1] < dep[t2])
swap(t1, t2), swap(v, u);
ans = max(ans, query(id[t1], id[v], 1));
v = fat[t1], t1 = top[v];
}
if(v == u) return ans;
if(dep[v] > dep[u]) swap(v, u);
return max(ans, query(id[son[v]], id[u], 1));
}
void slove()
{
char str[20];
int a, b;
while(scanf(" %s", str), str[0] != 'D')
{
scanf("%d%d", &a, &b);
if(str[0] == 'C')
update_edge(id[d[a][1]], b, 1);
else if(str[0] == 'N')
renew(a, b);
else if(str[0] == 'Q')
printf("%d\n", seek(a, b));
}
}
int main()
{
int t;
scanf("%d", &t);
while(t--)
{
num = cnt = 0;
memset(head, -1, sizeof head);
scanf("%d", &n);
for(int i = 1; i <= n - 1; i++)
{
scanf("%d%d%d", &d[i][0], &d[i][1], &d[i][2]);
add_edge(d[i][0], d[i][1]);
add_edge(d[i][1], d[i][0]);
}
dfs1(1, 0, 1);
dfs2(1, 1);
for(int i = 1; i <= n - 1; i++)
{
if(dep[d[i][0]] > dep[d[i][1]]) swap(d[i][0], d[i][1]);
val[id[d[i][1]]] = d[i][2];
}
build(1, num, 1);
slove();
}
return 0;
}