P2486 [SDOI2011]染色 - 树剖
先考虑在链上的做法,线段树子节点合并的时候减去重复算的颜色,然后树剖,跨越轻链的时候可以单点查询找颜色(我想的是维护颜色。。。但是明显这个点也是在线段树上的,直接单点查询就好了,想想问题的本质是什么,有没有不那么麻烦的做法)
所以要考虑好一些问题的区间可维护性,然后选用适当的数据结构
比如区间最大公约数,显然不同的区间可以合并维护,可以用线段树或者st表在logn时间内维护好,但是st表常数更小
容易写出错的地方:
应该写成左右子节点:query(w << 1 | 1, mid+1, r, x, y);
,然而w<<1|1写成了w。。。
树剖建树应该在两次dfs之后,并且用rnk数组映射dfn序为编号
路径查询/修改的时候比较的应该是两个点的dep[top[x]],比较的是深度!
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
using namespace std;
#define debug(x) cerr << #x << "=" << x << endl;
const int MAXN = 200000 + 10;
int n,m,last[MAXN],cnt_div,edge_tot,col[MAXN],dep[MAXN],siz[MAXN],dfn[MAXN];
int top[MAXN],rnk[MAXN],fa[MAXN],son[MAXN];
struct Edge{
int u, v, to;
Edge(){}
Edge(int u, int v, int to) : u(u), v(v), to(to) {}
}e[MAXN * 2];
inline void add(int u, int v) {
e[++edge_tot] = Edge(u, v, last[u]);
last[u] = edge_tot;
}
struct segment{
int sum, siz, coll, colr, add;
}tr[MAXN*4];
void update(int w) {
int le = w << 1, ri = w << 1 | 1;
tr[w].sum = tr[le].sum + tr[ri].sum;
tr[w].coll = tr[le].coll, tr[w].colr = tr[ri].colr;
if(tr[le].colr == tr[ri].coll) tr[w].sum--;
}
void down(int w) {
int le = w << 1, ri = w << 1 | 1, add = tr[w].add;
if(add == -1) return;
tr[w].add = -1;
tr[le].add = tr[le].coll = tr[le].colr = add;
tr[ri].add = tr[ri].coll = tr[ri].colr = add;
tr[le].sum = tr[ri].sum = 1;
tr[w].add = -1;
}
void build(int w, int l, int r) {
tr[w].add = -1;
if(l == r) {
tr[w].sum = 1;
tr[w].coll = col[rnk[l]], tr[w].colr = col[rnk[l]];
return;
}
int mid = (l + r) >> 1;
build(w<<1, l, mid);
build(w<<1|1, mid+1, r);
update(w);
}
int query(int w, int l, int r, int x, int y) {
if(x <= l && r <= y) {
return tr[w].sum;
}
down(w);
int mid = (l + r) >> 1;
int sum = 0;
int le = w << 1, ri = w << 1 | 1;
if(x <= mid) sum += query(le, l, mid, x, y);
if(y > mid) sum += query(ri, mid+1, r, x, y);
if(x <= mid && y > mid)
if(tr[le].colr == tr[ri].coll) sum--;
return sum;
}
int sin_que(int w, int l, int r, int pos) {
if(l == r) {
return tr[w].coll;
}
down(w);
int mid = (l + r) >> 1;
int le = w << 1, ri = w << 1 | 1;
if(pos <= mid) return sin_que(le, l, mid, pos);
return sin_que(ri, mid+1, r, pos);
}
void change(int w, int l, int r, int x, int y, int k) {
if(x <= l && r <= y) {
tr[w].add = tr[w].coll = tr[w].colr = k;
tr[w].sum = 1;
return;
}
down(w);
int mid = (l + r) >> 1;
int le = w << 1, ri = w << 1 | 1;
if(x <= mid) change(le, l, mid, x, y, k);
if(y > mid) change(ri, mid+1, r, x, y, k);
update(w);
}
void dfs_init(int x, int depth, int f) {
dep[x] = depth;
siz[x] = 1;
fa[x] = f;
for(int i=last[x]; i; i=e[i].to) {
int v = e[i].v;
if(v == f) continue;
dfs_init(v, depth+1, x);
if(siz[v] > siz[son[x]]) son[x] = v;
siz[x] += siz[v];
}
}
void dfs_top(int x, int t) {
top[x] = t;
dfn[x] = ++cnt_div;
rnk[cnt_div] = x;
if(son[x]) dfs_top(son[x], t);
for(int i=last[x]; i; i=e[i].to) {
int v = e[i].v;
if(v == fa[x] || v == son[x]) continue;
dfs_top(v, v);
}
}
int query_path(int s, int t) {
int sum = 0;
while(top[s] != top[t]) {
if(dep[top[s]] < dep[top[t]]) swap(s, t); //这里十分容易忘了写dep[]
sum += query(1, 1, n, dfn[top[s]], dfn[s]);
if(sin_que(1, 1, n, dfn[top[s]]) == sin_que(1, 1, n, dfn[fa[top[s]]])) sum--;
s = fa[top[s]];
}
if(dfn[s] > dfn[t]) swap(s, t);
sum += query(1, 1, n, dfn[s], dfn[t]);
return sum;
}
void change_path(int s, int t, int k) {
while(top[s] != top[t]) {
if(dep[top[s]] < dep[top[t]]) swap(s, t);
change(1, 1, n, dfn[top[s]], dfn[s], k);
s = fa[top[s]];
}
if(dfn[s] > dfn[t]) swap(s, t);
change(1, 1, n, dfn[s], dfn[t], k);
}
int main() {
scanf("%d%d", &n, &m);
for(int i=1; i<=n; i++) {
scanf("%d", &col[i]);
}
for(int i=1; i<n; i++) {
int a, b;
scanf("%d%d", &a, &b);
add(a, b);
add(b, a);
}
dfs_init(1, 1, 0);
dfs_top(1, 1);
build(1, 1, n);
char temp_que[5];
for(int i=1; i<=m; i++) {
scanf("%s", temp_que);
int s, t, c;
if(temp_que[0] == 'C') {
scanf("%d%d%d", &s, &t, &c);
change_path(s, t, c);
} else {
scanf("%d%d", &s, &t);
printf("%d\n", query_path(s, t));
}
}
return 0;
}