【CodeChef-ANUDTQ】Dynamic Trees and Queries【Splay】【DFS序】

【题目链接】

第一次见到用Splay维护DFS序的题。

（其实一开始看jiry_2题解，写的LCT，结果TLE / RE。重新思考人生后换成了Splay）

又忘把标记下传了。。

/* Pigonometry */
#include <cstdio>
#include <algorithm>

using namespace std;

typedef long long LL;

const int maxn = 200005, maxm = 100005, maxnode = 400005;

int n, m, head[maxn], cnt, q[maxn], tot, val[maxn];

struct _edge {
	int v, next;
} g[maxm << 1];
	
template <class numtype>
inline void read(numtype &x) {
	bool f = 0; x = 0; char ch = getchar();
	for(; ch < '0' || ch > '9'; ch = getchar()) f = (ch == '-');
	for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
	if(f) x = -x;
}

inline void add(int u, int v) {
	g[cnt] = (_edge){v, head[u]};
	head[u] = cnt++;
}

inline void dfs(int x, int f) {
	q[++tot] = (x << 1) - 1;
	for(int i = head[x]; ~i; i = g[i].next) if(g[i].v ^ f)
		dfs(g[i].v, x);
	q[++tot] = x << 1;
}

/* Splay */
int son[maxnode][2], size[maxnode], pre[maxnode], root, w[maxnode], addv[maxnode];
LL sum[maxnode];

inline void pushup(int x) {
	int l = son[x][0], r = son[x][1];
	size[x] = size[l] + 1 + size[r];
	sum[x] = sum[l] + w[x] + sum[r];
}

inline void pushdown(int x) {
	int l = son[x][0], r = son[x][1];
	if(addv[x]) {
		if(l) {
			w[l] += addv[x]; 
			sum[l] += (LL)size[l] * addv[x];
			addv[l] += addv[x];
		}
		if(r) {
			w[r] += addv[x];
			sum[r] += (LL)size[r] * addv[x];
			addv[r] += addv[x];
		}
		addv[x] = 0;
	}
}

inline void newnode(int &x, int f, int c) {
	x = ++tot;
	pre[x] = f;
	w[x] = sum[x] = c;
	addv[x] = 0;
}

inline int build(int l, int r, int f) {
	if(l > r) return 0;
	int mid = l + r >> 1, x = q[mid];
	pre[x] = f;
	w[x] = sum[x] = val[(x + 1) >> 1];
	addv[x] = 0;
	son[x][0] = build(l, mid - 1, x); son[x][1] = build(mid + 1, r, x);
	pushup(x);
	return x;
}	

inline void init() {
	root = 0;
	son[0][0] = son[0][1] = size[0] = pre[0] = sum[0] = 0;
	root = build(1, tot, root);
	pushup(root);
}

inline void rotate(int x) {
	int y = pre[x], z = pre[y], type = son[y][1] == x;
	pushdown(y); pushdown(x);
	pre[son[y][type] = son[x][!type]] = y;
	pre[x] = z;
	if(z) son[z][son[z][1] == y] = x;
	pre[son[x][!type] = y] = x;
	pushup(y); pushup(x);
}

inline void splay(int x, int goal) {
	while(pre[x] ^ goal) {
		int y = pre[x], z = pre[y];
		if(z == goal) rotate(x);
		else if(son[z][1] == y ^ son[y][1] == x) rotate(x), rotate(x);
		else rotate(y), rotate(x);
	}
	if(!goal) root = x;
}

inline int getpre(int x) {
	for(x = son[x][0]; son[x][1]; x = son[x][1]);
	return x;
}

inline int getsuf(int x) {
	for(x = son[x][1]; son[x][0]; x = son[x][0]);
	return x;
}

inline void insert(int x, int c) {
	x = (x << 1) - 1; splay(x, 0);
	int y = getsuf(x); splay(y, x);
	newnode(son[y][0], y, c);
	newnode(son[son[y][0]][1], son[y][0], c);
	pushup(son[son[y][0]][1]); pushup(son[y][0]);
	pushup(y); pushup(x);
}

inline void addw(int x, int c) {
	x = (x << 1) - 1;
	splay(x, 0); splay(x + 1, x);
	int y = son[x + 1][0];
	if(y) {
		w[y] += c;
		addv[y] += c;
		sum[y] += (LL)size[y] * c;
	}
	w[x] += c; w[x + 1] += c;
	pushup(x + 1); pushup(x);
}

inline void del(int x) {
	x = (x << 1) - 1;
	splay(x, 0); splay(x + 1, x);
	int l = getpre(x), r = getsuf(x + 1);
	splay(l, 0); splay(r, l);
	pre[son[r][0]] = 0; son[r][0] = 0;
	pushup(r); pushup(l);
}

inline LL query(int x) {
	x = (x << 1) - 1; 
	splay(x, 0); splay(x + 1, x);
	return (sum[son[x + 1][0]] >> 1) + w[x];
}

int main() {
	read(n);
	for(int i = 1; i <= n; i++) read(val[i]), head[i] = -1; cnt = tot = 0;
	for(int i = 1; i < n; i++) {
		int u, v; read(u); read(v); u++; v++;
		add(u, v); add(v, u);
	}

	dfs(1, 0);
	init();

	LL ans = 0;
	read(m);
	while(m--) {
		int opt; read(opt);
		LL x; read(x); x += ans + 1;
		if(opt == 1) {
			int v; read(v);
			insert(x, v);
		}
		else if(opt == 2) {
			int v; read(v);
			addw(x, v);
		}
		else if(opt == 3)
			del(x);
		else if(opt == 4)
			printf("%lld\n", ans = query(x));
	}
	return 0;
}

附比样例强一点的数据

2
10 20
0 1
9
4 0
1 -30 5
2 -30 1
4 -30
3 -37
4 -38
1 -15 233
4 -17
4 -248