题外话
首先这是个挺裸的题，由于太久没写剖分导致调了好久，前天调了一下午，一直查不到错
昨晚在看春晚的时候突然灵机一动，发现合并的时候出了问题，开电脑把它A掉了= =
感觉自己也蛮拼的
Description
给定一棵有n个节点的无根树和m个操作，操作有2类：
1:将节点a到节点b路径上所有点都染成颜色c
2:询问节点a到节点b路径上的颜色段数量(连续相同颜色被认为是同一段)，如"112221"由3段组成:"11"、"222"和"1"
Solution
很显然这是要剖分的辣，然后用线段树维护信息。然后我们会发现合并的时候有点蛋疼
它要求颜色段数量，所以我们很自然的想到要维护左右端的颜色信息来进行合并
线段树合并时sum[rt]=sum[ls]+sum[rs]?(rc[ls]==lc[rs])即可
但是我们正常在查询的时候是要在剖分后的树上”跳来跳去的”
所以每次查询的时候我们都需要维护当前左右端颜色信息，然后在”跳”的过程中合并
注意颜色是[0,109]的，所以打标记时要开始赋值为?1
合并的时候想清楚，注意细节就好辣！可以仔细想想或者详情参见代码QvQ
Code
#include <bits/stdc++.h>//树链剖分
using namespace std;
#define ls (rt << 1)
#define rs (rt << 1 | 1)
const int N = 100005, M = N << 2;
int n, m, lcol, rcol, tot, cnt, w[N], q[N], top[N], sz[N], son[N], pre[N], id[N], dep[N], col[N], lc[M], rc[M], sum[M], mark[M], to[M], nxt[M], head[N];
bool vis[N];
inline int readInt() {
char c;
while (c = getchar(), c < ‘0‘ || c > ‘9‘);
int t = c - ‘0‘;
while (c = getchar(), c >= ‘0‘ && c <= ‘9‘) t = t * 10 + c - ‘0‘;
return t;
}
void add(int u, int v) {
to[tot] = v, nxt[tot] = head[u], head[u] = tot++;
to[tot] = u, nxt[tot] = head[v], head[v] = tot++;
}
void pushdown(int rt) {
if (~mark[rt]) {
lc[ls] = lc[rs] = rc[ls] = rc[rs] = mark[ls] = mark[rs] = mark[rt];
sum[ls] = sum[rs] = 1;
mark[rt] = -1;
}
}
void pushup(int rt) {
lc[rt] = lc[ls], rc[rt] = rc[rs];
sum[rt] = sum[ls] + sum[rs] - (rc[ls] == lc[rs]);
}
void build(int rt, int l, int r) {
mark[rt] = -1;
if (l == r) {
lc[rt] = rc[rt] = w[l];
sum[rt] = 1;
return ;
}
int mid = l + r >> 1;
build(ls, l, mid);
build(rs, mid + 1, r);
pushup(rt);
}
void change(int rt, int l, int r, int L, int R, int c) {
if (L <= l && R >= r) {
lc[rt] = rc[rt] = mark[rt] = c;
sum[rt] = 1;
return;
}
pushdown(rt);
int mid = l + r >> 1;
if (L <= mid) change(ls, l, mid, L, R, c);
if (R > mid) change(rs, mid + 1, r, L, R, c);
pushup(rt);
}
int ask(int rt, int l, int r, int L, int R) {
if (L == l) lcol = lc[rt];
if (R == r) rcol = rc[rt];
if (L <= l && R >= r) return sum[rt];
pushdown(rt);
int mid = l + r >> 1, t = 0, t1 = -1, t2 = -1;
if (L <= mid) t += ask(ls, l, mid, L, R), t1 = rc[ls];
if (R > mid) t += ask(rs, mid + 1, r, L, R), t2 = lc[rs];
t -= (t1 == t2 && ~t1);
return t;
}
void work(int a, int b, int c) {
while (top[a] != top[b]) {
if (dep[top[a]] < dep[top[b]]) swap(a, b);
change(1, 1, n, id[top[a]], id[a], c);
a = pre[top[a]];
}
if (dep[a] < dep[b]) swap(a, b);
change(1, 1, n, id[b], id[a], c);
}
int work2(int a, int b) {
int t = 0, t1 = -1, t2 = -1;
while (top[a] != top[b]) {
if (dep[top[a]] < dep[top[b]]) swap(a, b), swap(t1, t2);
t += ask(1, 1, n, id[top[a]], id[a]);
t -= (t1 == rcol && ~t1);
t1 = lcol;
a = pre[top[a]];
}
if (dep[a] < dep[b]) swap(a, b), swap(t1, t2);
t += ask(1, 1, n, id[b], id[a]);
t -= ((t1 == rcol && ~t1) + (t2 == lcol && ~t2));
return t;
}
void gao() {
int r = 0;
vis[dep[1] = q[0] = 1] = 1;
for (int i = 0; i <= r; ++i)
for (int j = head[q[i]]; ~j; j = nxt[j])
if (!vis[to[j]]){
vis[to[j]] = 1;
dep[q[++r] = to[j]] = dep[q[i]] + 1;
pre[q[r]] = q[i];
}
for (int i = r; i >= 0; --i) {
sz[pre[q[i]]] += ++sz[q[i]];
if (sz[son[pre[q[i]]]] < sz[q[i]]) son[pre[q[i]]] = q[i];
}
for (int i = 0; i <= r; ++i)
if (!top[q[i]]) {
for (int j = q[i]; j; j = son[j]) {
top[j] = q[i];
w[id[j] = ++cnt] = col[j];
}
}
build(1, 1, n);
}
int main() {
n = readInt(), m = readInt();
memset(head, -1, sizeof(head));
for (int i = 1; i <= n; ++i) col[i] = readInt();
for (int i = 1, x, y; i < n; ++i) {
x = readInt(), y = readInt();
add(x, y);
}
gao();
while (m--) {
char s[5];
scanf("%s", s);
int a, b, c;
if (s[0] == ‘C‘) {
a = readInt(), b = readInt(), c = readInt();
work(a, b, c);
}
else {
a = readInt(), b = readInt();
printf("%d\n", work2(a, b));
}
}
return 0;
}
时间： 2024-08-02 07:02:22