題目分析
其他人要麼倍增，要麼左偏樹，那我就來講講樸實無華的 dfs 序加上線段樹的做法。
首先發現題目中明確指出了作乘法的時候一定是乘上一個大於零的數，這是為什麼呢？首先把可以佔領當前城池的戰鬥力的不等式列出來：
\[h_j \le \left\{
\begin{array}{c}
s_i \times v_j & & {a_j = 1}\\
s_i + v_j & & {a_j=0}
\end{array}
\right.
\]
發現當 \(v_j > 0\) 時，不等式不會變號，得到如下式子：
\[s_i \ge \left\{
\begin{array}{c}
\cfrac{h_j}{v_j} & & {a_j = 1}\\
h_j - v_j & & {a_j=0}
\end{array}
\right.
\]
於是，我們發現，判斷一大堆騎士能不能佔領只用看其中的最小血量是否滿足要求就行了。但是程式碼實現的時候為了避免丟失精度，不使用浮點數比較。對於當前城池，我們要知道起子樹內所有能跳到這裡來的騎士的最小血量，刪去犧牲在這裡的騎士，將留下來的騎士血量按照題意操作，在往上跳一步。子樹的問題可以用 dfs 序轉變成一個區間上的問題，區間最小值、區間乘、區間加、單點刪除，明顯可以用線段樹維護。刪除的時候將其戰鬥力設為 \(\infty\) 就不會對之後的造成影響。時間複雜度 \(\Theta(n \log n)\)，令 \(n\) 和 \(m\) 同階。
實際程式碼實現起來碼量很大？細節需要處理到位，特別是這題 dfs 序記的不是結點，而是士兵，所以會略有不同，具體見程式碼。
程式碼（已略去快讀快寫，碼風清新，註釋詳盡）
dfs 序
//#pragma GCC optimize(3)
//#pragma GCC optimize("Ofast", "inline", "-ffast-math")
//#pragma GCC target("avx", "sse2", "sse3", "sse4", "mmx")
#include <iostream>
#include <cstdio>
#define debug(a) cerr << "Line: " << __LINE__ << " " << #a << endl
#define print(a) cerr << #a << "=" << (a) << endl
#define file(a) freopen(#a".in", "r", stdin), freopen(#a".out", "w", stdout)
#define main Main(); signed main(){ return ios::sync_with_stdio(0), cin.tie(0), Main(); } signed Main
using namespace std;
#include <vector>
const int N = 300010;
const long long inf = 0x3f3f3f3f3f3f3f3fll;
int n, m;
vector<int> edge[N], man[N];
int ans1[N], ans2[N];
// 對於 ans2，轉變成初始位置深度 - 死亡位置深度，根節點深度 1，沒死的當做在深度為 0 的地方死了
int op[N], dpt[N];
long long h[N], v[N], s[N];
int L[N], R[N], val[N], timer;
void dfs(int now){
L[now] = timer + 1;
for (auto x: man[now]) val[++timer] = x;
for (auto to: edge[now]) dfs(to);
R[now] = timer;
}
// dfs 序記錄子樹所有士兵
struct Segment_Tree{
#define lson (idx << 1 )
#define rson (idx << 1 | 1)
struct Tag{
long long mul, add;
Tag operator + (const Tag & o) const {
return {mul * o.mul, add * o.mul + o.add};
}
inline void clear(){
mul = 1, add = 0;
}
};
// 懶惰標記
struct Info{
long long minn;
int pos;
Info operator + (const Info & o) const {
if (minn == inf) return o;
if (o.minn == inf) return *this;
if (minn < o.minn) return *this;
return o;
}
Info operator + (const Tag & o) const {
if (minn == inf) return *this;
return {minn * o.mul + o.add, pos};
}
};
// 資訊
struct node{
int l, r;
Info info;
Tag tag;
} tree[N << 2];
void pushup(int idx){
tree[idx].info = tree[lson].info + tree[rson].info;
}
void build(int idx, int l, int r){
tree[idx] = {l, r, inf, -1, 1, 0};
if (l == r) return tree[idx].info = {s[val[l]], l}, void();
int mid = (l + r) >> 1;
build(lson, l, mid), build(rson, mid + 1, r), pushup(idx);
}
void pushtag(int idx, const Tag t){
tree[idx].info = tree[idx].info + t;
tree[idx].tag = tree[idx].tag + t;
}
void pushdown(int idx){
pushtag(lson, tree[idx].tag), pushtag(rson, tree[idx].tag);
tree[idx].tag.clear();
}
Info query(int idx, int l, int r){
if (tree[idx].l > r || tree[idx].r < l) return {inf, -1};
if (l <= tree[idx].l && tree[idx].r <= r) return tree[idx].info;
return pushdown(idx), query(lson, l, r) + query(rson, l, r);
}
void modify(int idx, int l, int r, const Tag t){
if (tree[idx].l > r || tree[idx].r < l) return;
if (l <= tree[idx].l && tree[idx].r <= r) return pushtag(idx, t);
pushdown(idx), modify(lson, l, r, t), modify(rson, l, r, t), pushup(idx);
}
void erase(int pos){
modify(1, pos, pos, {0, inf});
}
void add(int l, int r, long long v){
modify(1, l, r, {1, v});
}
void mul(int l, int r, long long v){
modify(1, l, r, {v, 0});
}
void output(int idx){
if (tree[idx].l == tree[idx].r){
cerr << (tree[idx].info.minn == inf ? -1 : tree[idx].info.minn) << " \n"[tree[idx].l == timer];
return;
}
pushdown(idx), output(lson), output(rson);
}
#undef lson
#undef rson
} yzh;
// 貌似就是線段樹 2 ？
void redfs(int now){
if (L[now] > R[now]) return;
for (auto to: edge[now]) redfs(to);
// yzh.output(1);
while (true){
// 不斷刪去死了計程車兵，注意到士兵最多刪 m 次，故不會超時
Segment_Tree::Info res = yzh.query(1, L[now], R[now]);
if (res.pos == -1 || res.minn == inf) break;
if (res.minn >= h[now]) break;
ans2[val[res.pos]] -= dpt[now], yzh.erase(res.pos), ++ans1[now];
// yzh.output(1);
}
if (op[now]) yzh.mul(L[now], R[now], v[now]);
else yzh.add(L[now], R[now], v[now]);
}
// 第二次深搜求得答案
signed main(){
dpt[1] = 1, read(n, m);
for (int i = 1; i <= n; ++i) read(h[i]);
for (int i = 2, fa; i <= n; ++i) read(fa, op[i], v[i]), edge[fa].push_back(i), dpt[i] = dpt[fa] + 1;
for (int i = 1, pos; i <= m; ++i) read(s[i], pos), man[pos].push_back(i), ans2[i] = dpt[pos];
dfs(1), yzh.build(1, 1, timer), redfs(1);
for (int i = 1; i <= n; ++i) write(ans1[i], '\n');
for (int i = 1; i <= m; ++i) write(ans2[i], '\n');
return 0;
}
倍增（雖然沒講，但是也給出吧？）
//#pragma GCC optimize(3)
//#pragma GCC optimize("Ofast", "inline", "-ffast-math")
//#pragma GCC target("avx", "sse2", "sse3", "sse4", "mmx")
#include <iostream>
#include <cstdio>
#define debug(a) cerr << "Line: " << __LINE__ << " " << #a << endl
#define print(a) cerr << #a << "=" << (a) << endl
#define file(a) freopen(#a".in", "r", stdin), freopen(#a".out", "w", stdout)
#define main Main(); signed main(){ return ios::sync_with_stdio(0), cin.tie(0), Main(); } signed Main
using namespace std;
int n, m;
int op[300010];
int ans1[300010], ans2[300010];
int yzh[300010][20];
long long add[300010][20], mul[300010][20];
long long L[300010][20];
signed main(){
read(n, m);
for (int i = 1; i <= n; ++i) read(L[i][0]);
for (int i = 2, op; i <= n; ++i){
read(yzh[i][0], op), read(op ? mul[i][0] : (mul[i][0] = 1, add[i][0]));
}
for (int k = 1; k <= 19; ++k)
for (int i = 1; i <= n; ++i) if (!!(yzh[i][k] = yzh[yzh[i][k - 1]][k - 1])){
mul[i][k] = mul[i][k - 1] * mul[yzh[i][k - 1]][k - 1];
add[i][k] = add[i][k - 1] * mul[yzh[i][k - 1]][k - 1] + add[yzh[i][k - 1]][k - 1];
L[i][k] = max(L[i][k - 1], (L[yzh[i][k - 1]][k - 1] - add[i][k - 1] - 1) / mul[i][k - 1] + 1);
}
for (int i = 1, now; i <= m; ++i){
long long val; read(val, now);
for (int j = 19; j >= 0; --j)
if (yzh[now][j] && L[now][j] <= val)
ans2[i] += 1 << j, val = val * mul[now][j] + add[now][j], now = yzh[now][j];
if (val >= L[now][0]) ++ans2[i];
else ++ans1[now];
}
for (int i = 1; i <= n; ++i) write(ans1[i], '\n');
for (int i = 1; i <= m; ++i) write(ans2[i], '\n');
return 0;
}
左偏樹
//#pragma GCC optimize(3)
//#pragma GCC optimize("Ofast", "inline", "-ffast-math")
//#pragma GCC target("avx", "sse2", "sse3", "sse4", "mmx")
#include <iostream>
#include <cstdio>
#define debug(a) cerr << "Line: " << __LINE__ << " " << #a << endl
#define print(a) cerr << #a << "=" << (a) << endl
#define file(a) freopen(#a".in", "r", stdin), freopen(#a".out", "w", stdout)
#define main Main(); signed main(){ return ios::sync_with_stdio(0), cin.tie(0), Main(); } signed Main
using namespace std;
int n, m;
typedef int array[300010];
typedef long long Array[300010];
array lson, rson, root, a, dpt, fa, ans1, ans2, dis;
Array add, mul, h, s, v;
inline void pushtag(int x, long long mul, long long add){
::add[x] = ::add[x] * mul + add, ::mul[x] *= mul;
s[x] = s[x] * mul + add;
}
inline void pushdown(int x){
if (lson[x]) pushtag(lson[x], mul[x], add[x]);
if (rson[x]) pushtag(rson[x], mul[x], add[x]);
add[x] = 0, mul[x] = 1;
}
int merge(int x, int y){
if (!x || !y) return x | y;
if (s[x] > s[y]) swap(x, y);
pushdown(x), rson[x] = merge(rson[x], y);
if (dis[lson[x]] < dis[rson[x]]) swap(lson[x], rson[x]);
return dis[x] = dis[rson[x]] + 1, x;
}
signed main(){
dpt[1] = 1, read(n, m);
for (int i = 1; i <= n; ++i) read(h[i]);
for (int i = 2; i <= n; ++i) read(fa[i], a[i], v[i]), dpt[i] = dpt[fa[i]] + 1, mul[i] = 1;
for (int i = 1, bl; i <= m; ++i) read(s[i], bl), root[bl] = merge(root[bl], i), ans2[i] = dpt[bl];
for (int i = n; i >= 1; --i){
while (root[i] && s[root[i]] < h[i]){
ans2[root[i]] -= dpt[i], pushdown(root[i]), ++ans1[i];
root[i] = merge(lson[root[i]], rson[root[i]]);
}
if (i == 1) break;
if (root[i] == 0) continue;
if (a[i]) pushtag(root[i], v[i], 0);
else pushtag(root[i], 1, v[i]);
pushdown(root[i]), root[fa[i]] = merge(root[fa[i]], root[i]);
}
for (int i = 1; i <= n; ++i) write(ans1[i], '\n');
for (int i = 1; i <= m; ++i) write(ans2[i], '\n');
return 0;
}
總結 & 後話
線段樹無敵愛敲，另外兩種做法碼量小，速度快，雖然線段樹碼量大，速度慢，但是思路簡單是個不錯的選擇！