Description
You are given a tree with N nodes. The tree’s nodes are numbered 1 through N and its edges are numbered 1 through N ? 1. Each edge is associated with a weight. Then you are to execute a series of instructions on the tree. The instructions can be one of the following forms:
CHANGE i v : Change the weight of the ith edge to v
NEGATE a b : Negate the weight of every edge on the path from a to b
QUERY a b: Find the maximum weight of edges on the path from a to b
Input
The input contains multiple test cases. The first line of input contains an integer t (t ≤ 20), the number of test cases. Then follow the test cases.
Each test case is preceded by an empty line. The first nonempty line of its contains N (N ≤ 10,000). The next N ? 1 lines each contains three integers a, b and c, describing an edge connecting nodes a and b with weight c. The edges are numbered in the order they appear in the input. Below them are the instructions, each sticking to the specification above. A lines with the word “DONE” ends the test case.
Output
For each “QUERY” instruction, output the result on a separate line.
Sample Input
1 3 1 2 1 2 3 2 QUERY 1 2 CHANGE 1 3 QUERY 1 2 DONE
Sample Output
1 3
解析：很纯的一道树链剖分和线段树lazy标记的题，题解也有很多，这里只写下我处理的细节
细节：
1. 将边权赋给子节点，所以dfs求解父节点时，就直接给子节点赋予边的权值。同时为了之后的线段树的下标从1 ~ n-1根节点在树链剖分中的index 需要从0开始;
2. 对于改变某条边的权值，必须知道该边所对应的节点的id, 由于是链式建边的，所以最好使得2 ,3表示为第一条边，这样 i >> 1就表示当初输入时边的序号。
3. 线段树的pushdown操作注意下即可；（一是出现的位置不同，二是 是否及时pushdown)
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define lson l , m , rt<<1
#define rson m+1, r, rt<<1|1
#define inf 0x3f3f3f3f
#define MS1(a) memset(a,-1,sizeof(a))
const int maxn = 100007;
int head[maxn], tot, pos, son[maxn];
void init(){
memset(head, 0, sizeof(head));
pos = 0; tot = 1;
MS1(son);
}
struct edge{
int to, w, nxt;
}e[maxn<<1];
inline void ins(int u, int v,int w = 0)
{
e[++tot].nxt = head[u];
e[tot].to = v;
e[tot].w = w;
head[u] = tot;
}
int idx[maxn], weight[maxn];
int fa[maxn], cnt[maxn], dept[maxn];
void dfs(int u,int pre,int dep)
{
cnt[u] = 1;
fa[u] = pre;
dept[u] = dep;
for(int i = head[u]; i; i = e[i].nxt){
int v = e[i].to;
if(v != pre){
weight[v] = e[i].w; // 建线段树时，根据点来得到边权
idx[i>>1] = v; // change 边的id -> 子节点;
dfs(v, u, dep+1);
cnt[u] += cnt[v];
if(son[u] == -1 || cnt[son[u]] < cnt[v])
son[u] = v;
}
}
}
int p[maxn], fp[maxn], top[maxn];
void dfs(int u,int sp)
{
top[u] = sp;
p[u] = pos++; // pos++
fp[p[u]] = u;
if(son[u] == -1) return ;
dfs(son[u], sp);
for(int i = head[u]; i; i = e[i].nxt){
int v = e[i].to;
if(v != fa[u] && v != son[u])
dfs(v, v);
}
}
int mx[maxn], mn[maxn], lazy[maxn];
void pushup(int rt)
{
mx[rt] = max(mx[rt<<1], mx[rt<<1|1]);
mn[rt] = min(mn[rt<<1], mn[rt<<1|1]);
}
void build(int l, int r, int rt)
{
if(l == r){
mx[rt] = mn[rt] = weight[fp[l]]; // l为树链剖分之后点的id,需要转化为之前的id,得到权值;
return ;
}
int m = l + r >> 1;
if(l <= m) build(lson);
if(m < r) build(rson);
pushup(rt);
}
void print(int l,int r,int rt)
{
if(l == r){
printf("%d ",mx[rt]);
return ;
}
int m = l + r >> 1;
print(lson);
print(rson);
}
void pushdown(int rt)
{
if(lazy[rt]){
lazy[rt] = 0;
rt <<= 1;
int t = -mx[rt];
mx[rt] = -mn[rt];
mn[rt] = t;
lazy[rt] ^= 1;
rt |= 1;
t = -mx[rt];
mx[rt] = -mn[rt];
mn[rt] = t;
lazy[rt] ^= 1;
}
}
int query(int L, int R, int l,int r,int rt)
{
if(lazy[rt]) pushdown(rt);
if(L <= l && r <= R) return mx[rt];
int m = l + r >> 1;
int mx = -inf;
if(L <= m) mx = max(mx, query(L, R, lson));
if(m < R) mx = max(mx, query(L, R, rson));
return mx;
}
int n;
int query(int u,int v)
{
int fu = top[u], fv = top[v];
int ans = -inf;
while(fu != fv){
if(dept[fu] < dept[fv]){
swap(fu, fv); swap(u, v);
}
ans = max(ans, query(p[fu], p[u], 1, n-1, 1));
u = fa[fu];
fu = top[u];
}
if(u == v) return ans;
if(dept[u] < dept[v]) swap(u, v);
return max(ans, query(p[son[v]], p[u], 1, n-1, 1));
}
void update(int p,int val,int l,int r,int rt)
{
pushdown(rt);
if(l == r){
mx[rt] = mn[rt] = val;
return ;
}
int m = l + r >> 1;
if(p <= m) update(p, val, lson);
else update(p, val, rson);
pushup(rt);
}
void Change(int pos, int val)
{
int id = p[idx[pos]];
update(id, val, 1, n-1, 1);
}
void lazyNegate(int L,int R,int l,int r,int rt)
{
if(L <= l && r <= R){
int t = mx[rt];
mx[rt] = -mn[rt];
mn[rt] = -t;
lazy[rt] ^= 1;
return ;
}
pushdown(rt);
int m = l + r >> 1;
if(L <= m) lazyNegate(L, R, lson);
if(m < R) lazyNegate(L, R, rson);
pushup(rt);
}
void Negate(int u,int v)
{
int fu = top[u], fv = top[v];
while(fu != fv){
if(dept[fu] < dept[fv]){
swap(fu, fv); swap(u, v);
}
lazyNegate(p[fu], p[u], 1, n-1, 1);
u = fa[fu];
fu = top[u];
}
if(u == v) return ;
if(dept[u] < dept[v]) swap(u, v);
lazyNegate(p[son[v]], p[u], 1, n-1, 1);
}
int main()
{
//freopen("in.txt","r", stdin);
//freopen("out.txt","w",stdout);
int T;
cin >> T;
while(T--){
init();
int u, v, w;
scanf("%d", &n);
for(int i = 1;i < n; i++){
scanf("%d%d%d",&u, &v, &w);
ins(u,v,w);
ins(v,u,w);
}
dfs(1,0,0);
dfs(1,1);
build(1,n-1,1);
memset(lazy, 0, sizeof(lazy));
char op[10];
int a, b, cnt = 0;
while(scanf("%s", op) == 1, op[0] != ‘D‘){
scanf("%d%d", &a, &b);
if(op[0] == ‘Q‘)
printf("%d\n", query(a, b));
else if(op[0] == ‘C‘) Change(a, b);
else Negate(a, b);
//print(1,n-1,1);
}
}
}