Bzoj 2243(树链剖分+线段树)

BZOJ 2243
这题就是caioj 1102树上版。
所以我们树剖然后线段树维护

  • $lcol$:区间左端点的颜色
  • $rcol$:区间右端点的颜色
  • $cnt$:区间线段的条数

然后合并区间的时候如果中间颜色相同,要$cnt-1$
查询同理,如果左右区间都更新了,则要判断中间颜色

之后树剖的合并就比较麻烦。要分两边来存上一个区间端点颜色是什么,然后最后$u$到$v$的修改也要讨论,这个仔细想想就行,但是容易写错,建议画一下,详情看代码,不太好讲

注意树往深的方向是右方向,浅的方向是左方向,而且$pushdown$记得子节点要传$lazy$

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#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#define ms(i, j) memset(i, j, sizeof i)
#define LL long long
using namespace std;
const int MAXN = 100000 + 5;
int n, Q, col[MAXN], dep[MAXN], fa[MAXN], son[MAXN], siz[MAXN], top[MAXN], p[MAXN], np[MAXN], tb;
vector<int> G[MAXN];
void dfs1(int u, int pa) {
dep[u] = dep[pa] + 1, fa[u] = pa, siz[u] = 1;
for (int i = 0; i < (int)G[u].size(); i++) {
int v = G[u][i];
if (v != pa) {
dfs1(v, u);
siz[u] += siz[v];
if (son[u] == -1 || siz[v] > siz[son[u]]) son[u] = v;
}
}
}
void dfs2(int u, int chain) {
top[u] = chain;
p[u] = ++tb, np[p[u]] = u;
if (son[u] != -1) {
dfs2(son[u], chain);
for (int i = 0; i < (int)G[u].size(); i++) {
int v = G[u][i];
if (v != fa[u] && v != son[u]) {
dfs2(v, v);
}
}
}
}
#define lc (o << 1)
#define rc (o << 1 | 1)
#define M ((l + r) >> 1)
int lcol[MAXN * 4], rcol[MAXN * 4], cnt[MAXN * 4], lazy[MAXN * 4];
void pushup(int o, int l, int r) {
if (l == r) return ;
lcol[o] = lcol[lc], rcol[o] = rcol[rc], cnt[o] = cnt[rc] + cnt[lc];
if (rcol[lc] == lcol[rc]) cnt[o]--;
}
void pushdown(int o, int l, int r) {
if (l == r) return ;
if (lazy[o]) {
lazy[lc] = lazy[rc] = lazy[o];
lcol[lc] = lcol[rc] = lazy[o];
rcol[lc] = rcol[rc] = lazy[o];
cnt[lc] = cnt[rc] = 1;
lazy[o] = 0;
}
}
void build(int o, int l, int r) {
if (l == r) lcol[o] = rcol[o] = col[np[l]], cnt[o] = 1; else {
build(lc, l, M), build(rc, M + 1, r);
pushup(o, l, r);
}
}
void update(int o, int l, int r, int x, int y, int v) {
pushdown(o, l, r);
if (x <= l && r <= y) {
lazy[o] = v;
lcol[o] = rcol[o] = v, cnt[o] = 1;
return ;
}
if (x <= M) update(lc, l, M, x, y, v);
if (M < y) update(rc, M + 1, r, x, y, v);
pushup(o, l, r);
}
int nowl, nowr;
int query(int o, int l, int r, int x, int y) {
pushdown(o, l, r);
int ret = 0, flag = false;
if (x <= l && r <= y) {
if (l == x) nowl = lcol[o];
if (r == y) nowr = rcol[o];
return cnt[o];
}
if (x <= M) ret += query(lc, l, M, x, y), flag = true;
if (M < y) {
ret += query(rc, M + 1, r, x, y);
if (flag && rcol[lc] == lcol[rc]) ret--;
}
return ret;
}
void change(int u, int v, int c) {
int f1 = top[u], f2 = top[v];
while (f1 != f2) {
if (dep[f1] < dep[f2]) swap(f1, f2), swap(u, v);
update(1, 1, n, p[f1], p[u], c);
u = fa[f1], f1 = top[u];
}
if (dep[u] < dep[v]) swap(u, v);
update(1, 1, n, p[v], p[u], c);
}
int find(int u, int v) {
int ans = 0, f1 = top[u], f2 = top[v], last1l = -1, last1r = -1, last2l = -1, last2r = -1, isu = 1;
while (f1 != f2) {
if (dep[f1] < dep[f2]) swap(f1, f2), swap(u, v), swap(last1l, last2l), swap(last1r, last2r), isu = !isu;
nowl = nowr = -1;
int ret = query(1, 1, n, p[f1], p[u]);
if (last1l != -1 && nowr == last1l) ret--;
last1l = nowl, last1r = nowr, ans += ret;
u = fa[f1], f1 = top[u];
}
if (dep[u] < dep[v]) swap(u, v), swap(last1l, last2l), swap(last1r, last2r), isu = !isu;
nowl = nowr = -1;
int ret = query(1, 1, n, p[v], p[u]);
if (last1l != -1 && nowr == last1l) ret--;
if (last2l != -1 && nowl == last2l) ret--;
ans += ret;
return ans;
}
void clean() {
tb = 0;
for (int i = 0; i <= n * 4; i++) cnt[i] = lazy[i] = lcol[i] = rcol[i] = 0;
for (int i = 0; i <= n; i++) G[i].clear(), top[i] = p[i] = np[i] = dep[i] = fa[i] = 0, son[i] = -1;
}
void solve() {
clean();
for (int i = 1; i <= n; i++) scanf("%d", &col[i]);
for (int x, y, i = 1; i < n; i++) {
scanf("%d%d", &x, &y);
G[x].push_back(y), G[y].push_back(x);
}
dfs1(1, 0), dfs2(1, 1);
build(1, 1, n);
char s[10];
while (Q--) {
scanf("%s", s);
if (s[0] == 'C') {
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
change(a, b, c);
} else {
int a, b;
scanf("%d%d", &a, &b);
printf("%d\n", find(a, b));
}
}
}
int main() {
#ifndef ONLINE_JUDGE
freopen("1.in", "r", stdin);freopen("1.out", "w", stdout);
#endif
scanf("%d%d", &n, &Q), solve();
return 0;
}
/*
20 3
1 2 2 2 1 1 2 2 1 1 2 2 2 2 2 2 1 2 1 2
2 1
3 1
4 3
5 3
6 2
7 1
8 7
9 6
10 2
11 8
12 10
13 11
14 7
15 8
16 4
17 13
18 11
19 7
20 11
C 6 13 2
C 7 15 1
Q 12 13
*/
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