首先在子树里就是dfs序的一段区间。
那么路径最小值>=d的点呢?
很容易想到把点分树建出来,然后再上面×××
如果套上这个东西的话就变成了 O ( l o g 3 ) O(log^3) O(log3),还不说空间有多大。
这个其实就是kruskal重构树的事,模拟时sb了,没想到kruskal重构树可以套到这个上面。
满足路径最小值>=d的点同样在kruskal重构树的一个子树里,也用dfs序搞搞。
那么就是个三维偏序问题,考虑到这题卡空间,最好写的显然是KD-tree。
Code:
#include
#define fo(i, x, y) for(int i = x, B = y; i <= B; i ++)
#define ff(i, x, y) for(int i = x, B = y; i < B; i ++)
#define fd(i, x, y) for(int i = x, B = y; i >= B; i --)
#define ll long long
#define pp printf
#define hh pp("\n")
#define ul unsigned long long
using namespace std;
const int N = 1e5 + 5;
int n, m, ty, x, y, z, op;
ll a[N], la;
struct edge {
int fi[N * 2], nt[N * 2], to[N * 2], tot;
void link(int x, int y) {
nt[++ tot] = fi[x], to[tot] = y, fi[x] = tot;
}
int fa[N * 2], p[N * 2], q[N * 2], tp;
void dg(int x) {
p[x] = ++ tp;
for(int i = fi[x]; i; i = nt[i]) if(to[i] != fa[x])
fa[to[i]] = x, dg(to[i]);
q[x] = tp;
}
} e1, e2;
struct bb {
int x, y, z;
} b[N];
void init() {
scanf("%d %d %d", &n, &m, &ty);
fo(i, 1, n) scanf("%lld", &a[i]);
fo(i, 1, n - 1) {
scanf("%d %d %d", &x, &y, &z);
e1.link(x, y); e1.link(y, x);
b[i].x = x; b[i].y = y; b[i].z = z;
}
e1.dg(1);
}
int cmp(bb a, bb b) { return a.z > b.z;}
int f[N * 2], tf;
int F(int x) { return f[x] == x ? x : (f[x] = F(f[x]));}
int fa[18][N * 2], fv[18][N * 2];
void build_kruskal_tree() {
sort(b + 1, b + n, cmp);
fo(i, 1, 2 * n) f[i] = i;
tf = n;
fo(i, 1, n - 1) {
int x = F(b[i].x), y = F(b[i].y); z = b[i].z;
tf ++; f[x] = tf; f[y] = tf;
fa[0][x] = tf; fa[0][y] = tf;
fv[0][x] = z; fv[0][y] = z;
e2.link(tf, x); e2.link(tf, y);
}
e2.dg(tf);
fo(j, 1, 17) fo(i, 1, tf) {
fa[j][i] = fa[j - 1][fa[j - 1][i]];
fv[j][i] = fv[j - 1][fa[j - 1][i]];
}
}
int jump(int x, int y) {
fd(i, 17, 0) if(fa[i][x] && fv[i][x] >= y)
x = fa[i][x];
return x;
}
struct P {
int a[2], i;
} c[N];
int o, tt, rt, tc[N];
int cc(P a, P b) { return a.a[o] < b.a[o];}
struct tree {
int l, r;
int mi[2], mx[2], a[2];
ll lz, s;
} t[N];
#define i0 t[i].l
#define i1 t[i].r
void cmin(int &x, int y) { x > y ? x = y : 0;}
void cmax(int &x, int y) { x < y ? x = y : 0;}
void merge(tree &a, tree b) {
fo(j, 0, 1) cmin(a.mi[j], b.mi[j]), cmax(a.mx[j], b.mx[j]);
}
void bt(int &i, int x, int y) {
i = ++ tt;
int m = x + y >> 1;
nth_element(c + x, c + m, c + y + 1, cc);
fo(j, 0, 1) t[i].mi[j] = t[i].mx[j] = t[i].a[j] = c[m].a[j];
o = !o;
if(x < m) bt(i0, x, m - 1), merge(t[i], t[i0]);
if(m < y) bt(i1, m + 1, y), merge(t[i], t[i1]);
o = !o;
}
void build_kd_tree() {
fo(i, 1, n) c[i].a[0] = e1.p[i], c[i].a[1] = e2.p[i], c[i].i = i;
bt(rt, 1, n);
fo(i, 1, n) tc[c[i].i] = i;
}
void ad(int i, int x) {
t[i].lz += x, t[i].s += x;
}
void down(int i) {
if(t[i].lz) ad(i0, t[i].lz), ad(i1, t[i].lz), t[i].lz = 0;
}
int pl[2], pr[2], pw; ll px;
void add(int i, int x, int y) {
fo(j, 0, 1) if(t[i].mx[j] < pl[j] || t[i].mi[j] > pr[j]) return;
int ky = 1; fo(j, 0, 1) if(t[i].mi[j] < pl[j] || t[i].mx[j] > pr[j]) ky = 0;
if(ky) { ad(i, px); return;}
ky = 1; fo(j, 0, 1) if(t[i].a[j] < pl[j] || t[i].a[j] > pr[j]) ky = 0;
if(ky) t[i].s += px;
int m = x + y >> 1; down(i);
if(x < m) add(i0, x, m - 1);
if(m < y) add(i1, m + 1, y);
}
void ft(int i, int x, int y) {
int m = x + y >> 1; down(i);
if(m == pw) { px = t[i].s; return;}
if(pw < m) ft(i0, x, m - 1); else ft(i1, m + 1, y);
}
int main() {
freopen("ichi.in", "r", stdin);
freopen("ichi.out", "w", stdout);
init();
build_kruskal_tree();
build_kd_tree();
fo(i, 1, m) {
scanf("%d", &op);
if(op == 2) {
scanf("%d %d %d", &z, &y, &x);
if(ty) x = (x + la) % n + 1;
int nx = jump(x, y);
pl[0] = e1.p[x]; pr[0] = e1.q[x];
pl[1] = e2.p[nx]; pr[1] = e2.q[nx];
px = z;
add(rt, 1, n);
} else {
scanf("%d", &x);
if(ty) x = (x + la) % n + 1;
px = 0; pw = tc[x];
ft(1, 1, n);
la = a[x] + px;
pp("%lld\n", la);
}
}
}