线段树优化建图(cf787d, 2019Wannafly Winter Camp Day7 Div1 E)

线段树优化建图，用于区间到区间建边时降低空间复杂度
建立两颗线段树，一颗in, 代表进入这个区间，一颗out,代表从这个区间出去
in树从父亲向儿子建边，代表宏观进入整个区间，不向下寻找
out树从儿子向父亲建边，代表出去
in树向out树对应点建边，代表从这个点进去可以从它出去
建真正的边时：
1: 单点向单点: out树对应点向in树对应点建边
2: 单点向区间: out树对应点向in树对应区间建边
3: 区间向单点: out树对应区间向in树对应点建边
4: 区间向区间: out树区间对新点P建边，P向in树对应点建边

cf787D
最短路裸题

#include <bits/stdc++.h>
using namespace std;

const int N = 1e5 + 10;
using LL = long long;
vector<pair<int,int>> G[N << 3];
int in[N << 2], out[N << 2], pos[N], n, q, s, t, v, u, l, r, w, tot;
LL dis[N << 3];

inline void add(int x, int y, int v) {
	G[x].push_back(make_pair(y, v));
}

void build_in(int rt, int l, int r) {
	if(l == r) {
		pos[l] = rt;
		return;
	}
	int mid = l + r >> 1;
	build_in(rt << 1, l, mid);
	build_in(rt << 1 | 1, mid + 1, r);
	add(rt, rt << 1, 0); //in树父亲向儿子建边
	add(rt, rt << 1 | 1, 0);
}

void build_out(int rt, int l, int r) {
	add(rt, tot + rt, 0); //in树向out树建边
	if(l == r) {
		return;
	}
	int mid = l + r >> 1;
	build_out(rt << 1, l, mid);
	build_out(rt << 1 | 1, mid + 1, r);
	add(tot + (rt << 1), tot + rt, 0); //out树儿子向父亲建边
	add(tot + (rt << 1 | 1), tot + rt, 0);
}

void update_in(int rt, int l, int r, int L, int R, int from, int val) {
	if(L <= l && r <= R) {
		add(tot + pos[from], rt, val);
		return;
	}
	int mid = l + r >> 1;
	if(L <= mid)
		update_in(rt << 1, l, mid, L, R, from, val);
	if(mid < R)
		update_in(rt << 1 | 1, mid + 1, r, L, R, from, val);
}

void update_out(int rt, int l, int r, int L, int R, int pnt, int val) {
	if(L <= l && r <= R) {
		add(tot + rt, pos[pnt], val);
		return;
	}
	int mid = l + r >> 1;
	if(L <= mid)
		update_out(rt << 1, l, mid, L, R, pnt, val);
	if(mid < R)
		update_out(rt << 1 | 1, mid + 1, r, L, R, pnt, val);
}

struct node {
	LL dis;
	int id;
	bool operator<(const node &rhs) const {
		return dis > rhs.dis;
	}
};

void dijk() {
	memset(dis, 0x3f, sizeof(dis));
	dis[pos[s]] = 0;
	priority_queue<node> pq;
	pq.push({0, pos[s]});
	while(!pq.empty()) {
		node u = pq.top();
		pq.pop();
		if(dis[u.id] < u.dis) continue;
		for(auto &j: G[u.id]) {
			if(dis[j.first] > u.dis + j.second) {
				dis[j.first] = u.dis + j.second;
				pq.push({dis[j.first], j.first});
			}
		}
	}
}

int main() {
	scanf("%d%d%d", &n, &q, &s);
	tot = n << 2;
	build_in(1, 1, n);
	build_out(1, 1, n);
	while(q--) {
		scanf("%d", &t);
		if(t == 1) { //v->u
			scanf("%d%d%d", &v, &u, &w);
			add(pos[v] + tot, pos[u], w);
		}
		if(t == 2) { //v->[l,r]
			scanf("%d%d%d%d", &v, &l, &r, &w);
			update_in(1, 1, n, l, r, v, w);
		}
		if(t == 3) { //[l,r]->v
			scanf("%d%d%d%d", &v, &l, &r, &w);
			update_out(1, 1, n, l, r, v, w);
		}
	}
	dijk();
	for(int i = 1; i <= n; ++i) {
		printf("%lld%c", dis[pos[i]] == 0x3f3f3f3f3f3f3f3f ? -1 : dis[pos[i]], " 
"[i == n]);
	}

	return 0;
}

2019Wannafly Winter Camp Day7 Div1 E
给你线性探查法哈希后的序列，求字典序最小的原序列
记一个数应该在的位置为(pos), 实际在的位置为(s)，那么(pos)到(s-1)(模(n)意义下的)这些位置的数肯定在(s)前被插入
建边拓扑排序就行了，要求字典序最小就用优先队列，只有区间向单点建边只要out那颗树就行了

#include <bits/stdc++.h>
using namespace std;

const int N = 1e5 + 10;

vector<int> G[N << 2], ans;
int deg[N << 2], pos[N], id[N << 2], a[N], n;

void add(int x, int y) {
	G[x].push_back(y);
	++deg[y];
}

void build(int rt, int l, int r) {
	id[rt] = -1;
	if(l == r) {
		pos[l] = rt;
		id[rt] = l;
		return;
	}
	int mid = l + r >> 1;
	add(rt << 1, rt);
	add(rt << 1 | 1, rt);
	build(rt << 1, l, mid);
	build(rt << 1 | 1, mid + 1, r);
}

void Add(int rt, int l, int r, int L, int R, int pnt) {
	if(L <= l && r <= R) {
		add(rt, pos[pnt]);
		return;
	}
	int mid = l + r >> 1;
	if(L <= mid)
		Add(rt << 1, l, mid, L, R, pnt);
	if(mid < R)
		Add(rt << 1 | 1, mid + 1, r, L, R, pnt);
}

void topo() {
	priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;
	for(int i = 0; i < n; ++i) 
		if(!deg[pos[i]])
			pq.push(make_pair(a[i], pos[i]));
	while(!pq.empty()) {
		pair<int, int> u = pq.top();
		pq.pop();
		if(u.first != -1)
			ans.push_back(u.first);
		for(auto &it: G[u.second])
			if(--deg[it] == 0)
				pq.push(make_pair(~id[it] ? a[id[it]] : -1, it));
	}
	for(int i = 0; i < ans.size(); ++i) {
		printf("%d%c", ans[i], " 
"[i == ans.size() - 1]);
	}
}

int main() {
	scanf("%d", &n);
	build(1, 0, n - 1);
	for(int i = 0; i < n; ++i) {
		scanf("%d", &a[i]);
		int tmp = a[i] % n;
		if(tmp == i) continue;
		if(tmp < i)
			Add(1, 0, n - 1, tmp, i - 1, i);
		else {
			Add(1, 0, n - 1, tmp, n - 1, i);
			if(i)
				Add(1, 0, n - 1, 0, i - 1, i);
		}
	}
	topo();
	return 0;
}