1、题目大意：给个树，然后树上每个点都有颜色，然后会有路径的修改，有个询问，询问一条路径上的颜色分成了几段
2、分析：首先这个修改是树剖可以做的，对吧，但是这个分成了几段怎么搞呢，我们的树剖的不是要建线段树吗
我们的线段树存这样的几个值，一个是这个区间被分成了几段，另外就是这个区间的最左边的颜色和最右边的颜色
这样，我们在区间合并的时候把两个区间的段数加起来
然后用左区间的右端点和右区间的左端点如果相同就-1就可以了，那么这道题就做完了
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;
#define M 1000000
struct node{
int l, ans, r;
};
struct hehe{
int Top[M], Size[M], High[M], Fa[M], num[M], value[M];
node q[M];
int lazy[M];
int tot, head[M], Next[M], son[M];
int ST_tot;
int n;
inline void init(){
ST_tot = tot = 0;
for(int i = 1; i < M; i ++) lazy[i] = -1;
memset(head, -1, sizeof(head));
Top[1] = 1;
}
inline void pushdown(int o){
if(lazy[o] != -1){
lazy[2 * o] = lazy[2 * o + 1] = lazy[o];
q[2 * o].l = q[2 * o].r = q[2 * o + 1].l = q[2 * o + 1].r = lazy[o];
q[2 * o].ans = q[2 * o + 1].ans = 1;
lazy[o] = -1;
}
}
inline void add(int l, int r, int o, int x, int y, int k){
if(x <= l && r <= y){
q[o].l = q[o].r = k;
q[o].ans = 1;
lazy[o] = k;
return;
}
int mid = (l + r) / 2;
pushdown(o);
if(x <= mid) add(l, mid, 2 * o, x, y, k);
if(y > mid) add(mid + 1, r, 2 * o + 1, x, y, k);
q[o].l = q[2 * o].l;
q[o].r = q[2 * o + 1].r;
q[o].ans = q[2 * o].ans + q[2 * o + 1].ans;
if(q[2 * o].r == q[2 * o + 1].l) q[o].ans --;
}
inline node query(int l, int r, int o, int x, int y){
if(x <= l && r <= y){
return q[o];
}
pushdown(o);
int mid = (l + r) / 2;
if(x <= mid && y <= mid) return query(l, mid, 2 * o, x, y);
else if(x > mid && y > mid) return query(mid + 1, r, 2 * o + 1, x, y);
else{
node ll = query(l, mid, 2 * o, x, y);
node rr = query(mid + 1, r, 2 * o + 1, x, y);
node ret;
ret.l = ll.l;
ret.r = rr.r;
ret.ans = ll.ans + rr.ans;
if(ll.r == rr. l) ret.ans --;
return ret;
}
}
inline void insert(int x, int y){
tot ++;
son[tot] = y;
Next[tot] = head[x];
head[x] = tot;
}
inline void dfs1(int x, int fa, int height){
Fa[x] = fa;
High[x] = height;
for(int i = head[x]; i != -1; i = Next[i]) if(son[i] != fa){
dfs1(son[i], x, height + 1);
Size[x] += Size[son[i]];
}
Size[x] ++;
}
inline void dfs2(int x, int fa){
++ ST_tot;
num[x] = ST_tot;
add(1, n, 1, ST_tot, ST_tot, value[x]);
int o = -1, sos = 0;
for(int i = head[x]; i != -1; i = Next[i]) if(son[i] != fa){
if(Size[son[i]] > sos){
sos = Size[son[i]];
o = i;
}
}
if(o != -1){
Top[son[o]] = Top[x];
dfs2(son[o], x);
}
for(int i = head[x]; i != -1; i = Next[i]) if(son[i] != fa && i != o){
Top[son[i]] = son[i];
dfs2(son[i], x);
}
}
inline void real_add(int x, int y, int k){
while(Top[x] != Top[y]){
if(High[Top[x]] < High[Top[y]]) swap(x, y);
add(1, n, 1, num[Top[x]], num[x], k);
x = Fa[Top[x]];
}
if(High[x] < High[y]) swap(x, y);
add(1, n, 1, num[y], num[x], k);
return;
}
inline int real_query(int x, int y){
int tx = -1, ty = -1;
int ret = 0;
while(Top[x] != Top[y]){
if(High[Top[x]] < High[Top[y]]){
swap(x, y);
swap(tx, ty);
}
node hh = query(1, n, 1, num[Top[x]], num[x]);
ret += hh.ans;
if(hh.r == tx) ret --;
tx = hh.l;
x = Fa[Top[x]];
}
if(High[x] < High[y]){
swap(x, y);
swap(tx, ty);
}
node hh = query(1, n, 1, num[y], num[x]);
ret += hh.ans;
if(hh.r == tx) ret --;
if(hh.l == ty) ret --;
return ret;
}
} wt;
int main(){
int n, m;
scanf("%d%d", &n, &m);
wt.n = n;
wt.init();
for(int i = 1; i <= n; i ++) scanf("%d", &wt.value[i]);
for(int i = 1; i < n; i ++){
int x, y;
scanf("%d%d", &x, &y);
wt.insert(x, y);
wt.insert(y, x);
}
wt.dfs1(1, -1, 1);
wt.dfs2(1, -1);
char str[2];
int x, y, z;
for(int i = 1; i <= m; i ++){
scanf("%s", str);
if(str[0] == 'Q'){
scanf("%d%d", &x, &y);
printf("%d\n", wt.real_query(x, y));
}
else{
scanf("%d%d%d", &x, &y, &z);
wt.real_add(x, y, z);
}
}
return 0;
}