JZOJ 5460. 士兵训练

题目

(1le n，q le 2cdot {10}^5，0le b_i，l_i le {10}^9，b_i ge 1，1 le S_i le n)

(Solution)

这题很好想
总之要维护子树内 (b) 值的严格最大（包括数量），次大，次次大，(l) 值的严格最大，次大
然后分类讨论，注意相等的情况，接下来就是码的事了
注意要打人工栈！！

(Code)

#include<cstdio>
#include<iostream>
#define ls (p << 1)
#define rs (ls | 1)
using namespace std;

const int N = 2e5 + 5, INF = 2e9;
int n, q, dfc, tot, h[N], fa[N], dfn[N], rev[N], siz[N], B[N], L[N];
struct edge{int to, nxt;}e[N];
struct node{int mx, cmx, cnt, cc, mxl, cmxl;}seg[N << 2];

inline int add(int x, int y){e[++tot] = edge{y, h[x]}, h[x] = tot;}

int st[N], top; 
void dfs(int x)
{
	st[++top] = 1, dfn[x] = ++dfc, rev[dfc] = x, siz[x] = 1;
	while (top)
	{
		int x = st[top], v = e[h[x]].to;
		int bz = 0;
		if (h[x]){st[++top] = v, dfn[v] = ++dfc, rev[dfc] = v, siz[v] = 1, h[x] = e[h[x]].nxt, bz = 1;}
		if (!bz) top--, siz[fa[x]] += siz[x];
	}
}

inline node get1(node x, node y)
{
	if (x.mx == y.mx)
	{
		if (x.cmx == y.cmx) return node{x.mx, x.cmx, x.cnt + y.cnt, max(x.cc, y.cc)};
		else if (x.cmx > y.cmx) return node{x.mx, x.cmx, x.cnt + y.cnt, max(x.cc, y.cmx)};
		return node{x.mx, y.cmx, x.cnt + y.cnt, max(x.cmx, y.cc)};
	}
	else if (x.mx > y.mx) 
	{
		if (x.cmx == y.mx) return node{x.mx, x.cmx, x.cnt, max(x.cc, y.cmx)};
		else if (x.cmx > y.mx) return node{x.mx, x.cmx, x.cnt, max(x.cc, y.mx)};
		return node{x.mx, y.mx, x.cnt, max(x.cmx, y.cmx)};
	}
	else{
		if (x.mx == y.cmx) return node{y.mx, x.mx, y.cnt, max(x.cmx, y.cc)};
		else if (x.mx > y.cmx) return node{y.mx, x.mx, y.cnt, max(x.cmx, y.cmx)};
		return node{y.mx, y.cmx, y.cnt, max(x.mx, y.cc)};
	}
}
inline node get2(node x, node y)
{
	if (x.mxl == y.mxl) return node{0, 0, 0, 0, x.mxl, max(x.cmxl, y.cmxl)};
	else if (x.mxl > y.mxl) return node{0, 0, 0, 0, x.mxl, max(x.cmxl, y.mxl)};
	return node{0, 0, 0, 0, y.mxl, max(x.mxl, y.cmxl)};
}

void build(int l, int r, int p)
{
	if (l == r)
	{
		seg[p].mx = B[rev[l]], seg[p].mxl = L[rev[l]], seg[p].cnt = 1;
		seg[p].cmx = seg[p].cmxl = seg[p].cc = -INF;
		return;
	}
	int mid = (l + r) >> 1;
	build(l, mid, ls), build(mid + 1, r, rs);
	seg[p] = get1(seg[ls], seg[rs]);
	node K = get2(seg[ls], seg[rs]);
	seg[p].mxl = K.mxl, seg[p].cmxl = K.cmxl;
}
node query1(int l, int r, int p, int x, int y)
{
	if (x > y) return node{-INF, -INF, 0, -INF, -INF, -INF};
	if (x <= l && r <= y) return seg[p];
	int mid = (l + r) >> 1;
	node K1, K2;
	K1 = K2 = node{-INF, -INF, 0, -INF, -INF, -INF};
	if (x <= mid) K1 = query1(l, mid, ls, x, y);
	if (y > mid) K2 = query1(mid + 1, r, rs, x, y);
	return get1(K1, K2);
}
node query2(int l, int r, int p, int x, int y)
{
	if (x > y) return node{-INF, -INF, 0, -INF, -INF, -INF};
	if (x <= l && r <= y) return seg[p];
	int mid = (l + r) >> 1;
	node K1, K2, K;
	K1 = K2 = node{-INF, -INF, 0, -INF, -INF, -INF};
	if (x <= mid) K1 = query2(l, mid, ls, x, y);
	if (y > mid) K2 = query2(mid + 1, r, rs, x, y);
	return get2(K1, K2);
}

int main()
{
	freopen("soldier.in", "r", stdin);
	freopen("soldier.out", "w", stdout);
	scanf("%d%d", &n, &q);
	for(int i = 1; i < n; i++) scanf("%d", &fa[i + 1]), add(fa[i + 1], i + 1);
	for(int i = 1; i <= n; i++) scanf("%d%d", &B[i], &L[i]);
	dfs(1), build(1, n, 1);
	for(; q; --q)
	{
		int s, ans = 0; node K1, K2, K, Kl;
		scanf("%d", &s);
		K = query1(1, n, 1, dfn[s], dfn[s] + siz[s] - 1);
		if (siz[s] == 1){printf("0
"); continue;}
		K1 = query2(1, n, 1, 1, dfn[s] - 1), K2 = query2(1, n, 1, dfn[s] + siz[s], n), Kl = get2(K1, K2);
		if ((K.cnt > 1 && K.mx + Kl.mxl > K.mx) || (K.cmx + Kl.mxl > K.mx)) ans = K.mx;
		else{
			ans = K.cmx + Kl.mxl;
			if (ans == K.mx) ans = max(K.cmx + Kl.cmxl, K.cc + Kl.mxl);
			if (ans < 0) ans = K.cmx;
		}
		printf("%d
", ans);
	}
}