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http://www.lydsy.com/JudgeOnline/problem.php?id=3626
让我比较惊讶的一道链剖裸题(‘ ‘ ) 做法很精妙
首先我们考虑对于单个询问时可以拆分成(1, l - 1, z) 和 (1, r, z) 的, 然后考虑对于每一次询问可以表示为将(1, l) 的所有点到根的全部加1 然后求z到根路径的的和。 所以将询问离线, 按询问的l值排序,每一次遇到新的l值就将这一段的点到根的路径全部加1,然后查询即可
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long ll;
const ll maxn = 100001;
const ll mod = 201314;
ll n, m;
ll int_get() {
ll x = 0; char c = (char)getchar(); bool f = 0;
while(!isdigit(c)) {
if(c == ‘-‘) f = 1;
c = (char)getchar();
}
while(isdigit(c)) {
x = x * 10 + (int)(c - ‘0‘);
c = (char)getchar();
}
if(f) x = -x;
return x;
}
struct seg {
ll data, lazy;
seg *l, *r;
}tr[maxn * 3];ll sege = 0;
seg* root;
void test(seg* x, ll l, ll r) {
cout << l <<" "<< r <<" "<< x-> data <<" "<< x-> lazy << endl;
if(l ^ r) {
ll mid = (l + r) >> 1;
test(x-> l, l, mid), test(x-> r, mid + 1, r);
}
}
seg* build(ll l, ll r) {
seg* x = tr + sege ++;
if(l ^ r) {
ll mid = (l + r) >> 1;
x-> l = build(l, mid);
x-> r = build(mid + 1, r);
}
return x;
}
void update(seg* x) {
if(x-> l) x-> data = x-> l-> data + x-> r-> data;
}
void pushdown(seg* x, ll l, ll r) {
if(x-> l && x-> lazy != 0) {
ll mid = (l + r) >> 1;
x-> l-> lazy += x-> lazy, x-> l-> data += (mid - l + 1) * x-> lazy;
x-> r-> lazy += x-> lazy, x-> r-> data += (r - mid) * x-> lazy;
x-> lazy = 0;
}
}
void addlazy(seg* x, ll l, ll r, ll ls, ll rs, ll v) {
if(l == ls && r == rs) x-> data += v * (rs - ls + 1), x-> lazy += v;
else {
pushdown(x, l, r);
ll mid = (l + r) >> 1;
if(rs <= mid) addlazy(x-> l, l, mid, ls, rs, v);
else if(ls > mid) addlazy(x-> r, mid + 1, r, ls, rs, v);
else addlazy(x-> l, l, mid, ls, mid, v), addlazy(x-> r, mid + 1, r, mid + 1, rs, v);
update(x);
}
}
ll ask(seg* x, ll l, ll r, ll ls, ll rs) {
if(l == ls && r == rs) return x-> data;
else {
pushdown(x, l, r);
ll mid = (l + r) >> 1;
if(rs <= mid) return ask(x-> l, l, mid, ls, rs);
else if(ls > mid) return ask(x-> r, mid + 1, r, ls, rs);
else return ask(x-> l, l, mid, ls, mid) + ask(x-> r, mid + 1, r, mid + 1, rs);
}
}
struct edge {
ll t;
edge* next;
}e[maxn * 2], *head[maxn]; ll ne = 0;
void addedge(ll f, ll t) {
e[ne].t = t, e[ne].next = head[f], head[f] = e + ne ++;
}
ll h[maxn], size[maxn], fa[maxn], un[maxn], map[maxn], num = 0;
void dfs(ll x, ll pre) {
fa[x] = pre, size[x] = 1; h[x] = h[pre] + 1;
for(edge* p = head[x]; p; p = p-> next) {
if(p-> t != pre) dfs(p-> t, x), size[x] += size[p-> t];
}
}
void divide(ll x, ll Un) {
un[x] = Un, map[x] = ++ num;
if(size[x] == 1) return ;
ll Max = 0, pos;
for(edge* p = head[x]; p; p = p-> next) {
if(p-> t != fa[x] && size[p-> t] > Max) Max = size[p-> t], pos = p-> t;
}
divide(pos, Un);
for(edge* p = head[x]; p; p = p-> next) {
if(p-> t != fa[x] && p-> t != pos) divide(p-> t, p-> t);
}
}
void add(ll a, ll b, ll v) {
ll ls, rs;
while(un[a] != un[b]) {
if(h[un[a]] > h[un[b]]) {
ls = map[un[a]], rs = map[a];
addlazy(root, 1, n, ls, rs, v);
a = fa[un[a]];
}
else {
ls = map[un[b]], rs = map[b];
addlazy(root, 1, n, ls, rs, v);
b = fa[un[b]];
}
}
ls = map[a], rs = map[b];
if(ls > rs) swap(ls, rs);
addlazy(root, 1, n, ls, rs, v);
}
ll get(ll a, ll b) {
ll ret = 0;
ll ls, rs;
while(un[a] != un[b]) {
if(h[un[a]] > h[un[b]]) {
ls = map[un[a]], rs = map[a];
ret += ask(root, 1, n, ls, rs);
a = fa[un[a]];
}
else {
ls = map[un[b]], rs = map[b];
ret += ask(root, 1, n, ls, rs);
b = fa[un[b]];
}
}
ls = map[a], rs = map[b];
if(ls > rs) swap(ls, rs);
ret += ask(root, 1, n, ls, rs);
return ret;
}
struct block {
ll x, p, pos, fl;
}o[maxn]; ll oe = 0;
bool cmp(block a, block b) {
return a. x < b. x;
}
ll ans[maxn];
void read() {
n = int_get(); m = int_get();
for(ll i = 2; i <= n; ++ i) {
ll u = int_get() + 1;
addedge(u, i), addedge(i, u);
}
for(ll i = 1; i <= m; ++ i) {
ll a, b, c;
a = int_get() + 1, b = int_get() + 1, c = int_get() + 1;
++ oe, o[oe]. x = a - 1, o[oe]. p = c, o[oe]. pos = i, o[oe]. fl = -1;
++ oe, o[oe]. x = b, o[oe]. p = c, o[oe].pos = i, o[oe].fl = 1;
}
dfs(1, 0), divide(1, 1);
root = build(1, n);
}
void sov() {
sort(o + 1, o + 1 + oe, cmp); ll pl = 0;
for(ll i = 1; i <= oe; ++ i) {
while(pl < o[i]. x) {
++ pl; add(1, pl, 1);
}
//test(root, 1, n); cout << endl;
ans[o[i]. pos] += o[i]. fl * get(1, o[i]. p);
}
for(ll i = 1; i <= m; ++ i) printf("%lld\n", (ans[i] % mod + mod)% mod);
}
int main() {
//freopen("test.in", "r", stdin);
//freopen("test.out", "w", stdout);
read();
sov();
return 0;
}
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原文地址:http://www.cnblogs.com/ianaesthetic/p/4254722.html