1A!!! 哈哈哈哈哈没看题解 没套模板哈哈哈哈 太感动了!!
如果只是线段树的话这道题倒是不难,只要记录左右边界就好了,类似很久以前做的hotel的题
但是树上相邻的段会有连续的
树上top[x]和fa[top[x]]是连续的,但是线段树上是算不到的,所以要判断下
线段树记录的是区间的数量,但是求单点的时候求得是颜色,需要注意下
#include <iostream>
#include <cstring>
#include <cstdio> using namespace std;
const int N = ;
int a[N]; struct Edge {
int to, next;
} edge[N*];
int head[N], cntE; void addedge(int u, int v) {
edge[cntE].to = v; edge[cntE].next = head[u]; head[u] = cntE++;
edge[cntE].to = u; edge[cntE].next = head[v]; head[v] = cntE++;
} int dep[N], fa[N], sz[N], son[N];
void dfs1(int u, int pre, int d) {
dep[u] = d;
sz[u] = ;
fa[u] = pre;
for (int i = head[u]; ~i; i = edge[i].next) {
int v = edge[i].to;
if (v != pre) {
dfs1(v, u, d+);
sz[u] += sz[v];
if (son[u] == - || sz[v] > sz[son[u]]) son[u] = v;
}
}
} int top[N], dfn[N], rk[N], idx;
void dfs2(int u, int tp) {
top[u] = tp;
dfn[u] = ++idx;
rk[idx] = u;
if (son[u] == -) return ;
dfs2(son[u], tp);
for (int i = head[u]; ~i; i = edge[i].next) {
int v = edge[i].to;
if (v != fa[u] && v != son[u]) {
dfs2(v, v);
}
}
} int tr[N<<], ltr[N<<], rtr[N<<]; // tr[i] means the number of color segment
int fg[N<<];
#define lson o<<1
#define rson o<<1|1 void pushup(int o) {
ltr[o] = ltr[lson];
rtr[o] = rtr[rson];
tr[o] = tr[lson] + tr[rson];
if (rtr[lson] == ltr[rson]) tr[o]--;
} void pushdown(int o) {
if (fg[o]) {
tr[lson] = tr[rson] = ;
fg[lson] = fg[rson] = fg[o];
ltr[lson] = ltr[rson] = ltr[o];
rtr[lson] = rtr[rson] = rtr[o];
fg[o] = ;
}
} void build(int o, int l, int r) {
fg[o] = ;
if (l == r) {
tr[o] = ;
ltr[o] = rtr[o] = a[rk[l]];
return;
}
int mid = (l+r) >> ;
build(lson, l, mid);
build(rson, mid+, r);
pushup(o);
} void change(int o, int l, int r, int L, int R, int v) {
if (l >= L && r <= R) {
fg[o] = ;
tr[o] = ;
ltr[o] = rtr[o] = v;
return ;
}
pushdown(o);
int mid = (l+r) >> ;
if (mid >= L) change(lson, l, mid, L, R, v);
if (mid < R) change(rson, mid+, r, L, R, v);
pushup(o);
} void CHANGE(int x, int y, int n, int c) {
while (top[x] != top[y]) {
if (dep[top[x]] < dep[top[y]]) swap(x, y);
change(, , n, dfn[top[x]], dfn[x], c);
x = fa[top[x]];
}
if (dep[x] > dep[y]) swap(x, y);
change(, , n, dfn[x], dfn[y], c);
} int query(int o, int l, int r, int L, int R) {
if (l >= L && r <= R) return tr[o];
pushdown(o);
int mid = (l+r) >> ;
if (mid < L) {
return query(rson, mid+, r, L, R);
} else if (mid >= R) {
return query(lson, l, mid, L, R);
} else {
int ans = query(lson, l, mid, L, R);
ans += query(rson, mid+, r, L, R);
if (ltr[rson] == rtr[lson]) ans--;
return ans;
}
} int qq(int o, int l, int r, int p) {
if (l == r) return ltr[o];
pushdown(o);
int mid = (l+r) >> ;
if (mid >= p) return qq(lson, l, mid, p);
return qq(rson, mid+, r, p);
} int QUERY(int x, int y, int n) {
int ans = ;
while (top[x] != top[y]) {
if (dep[top[x]] < dep[top[y]]) swap(x, y);
ans += query(, , n, dfn[top[x]], dfn[x]);
if (qq(, , n, dfn[top[x]]) == qq(, , n, dfn[fa[top[x]]])) --ans;
x = fa[top[x]];
}
if (dep[x] > dep[y]) swap(x, y);
ans += query(, , n, dfn[x], dfn[y]);
return ans;
} void init() {
memset(head, -, sizeof head);
memset(son, -, sizeof son);
idx = cntE = ;
} int main()
{
//freopen("in.txt", "r", stdin);
int n, m;
while (~scanf("%d%d", &n, &m)) {
init();
for (int i = ; i <= n; ++i) scanf("%d", &a[i]);
int u, v, c;
for (int i = ; i < n; ++i) {
scanf("%d%d", &u, &v);
addedge(u, v);
}
dfs1(, , ); dfs2(, ); build(, , n); char op[];
while (m--) {
scanf("%s", op);
if (*op == 'Q') {
scanf("%d%d", &u, &v);
printf("%d\n", QUERY(u, v, n));
} else {
scanf("%d%d%d", &u, &v, &c);
CHANGE(u, v, n, c);
}
}
}
return ;
}