CodeForces 787D : Legacy 线段树优化建图 + 最短路

传送门

题意

分析

线段树优化建图的板子题
我们把区间映射到线段树的一个一个点上，然后建两颗线段树，第一颗线段树自上向下建边，边权为0，第二颗线段树自下往上建边，边权为0，这样就可以做到两颗线段树内的点互相走通
然后操作2可以在第一个线段树内加边，操作3可以在第二个线段树内加边

代码

#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define debug(x) cout<<#x<<":"<<x<<endl;
#define dl(x) printf("%lld\n",x);
#define di(x) printf("%d\n",x);
#define _CRT_SECURE_NO_WARNINGS
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, int> PII;
typedef vector<int> VI;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const int N = 2e6 + 10;
const ll mod = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
template<typename T>inline void read(T &a) {
	char c = getchar(); T x = 0, f = 1; while (!isdigit(c)) {if (c == '-')f = -1; c = getchar();}
	while (isdigit(c)) {x = (x << 1) + (x << 3) + c - '0'; c = getchar();} a = f * x;
}
int gcd(int a, int b) {return (b > 0) ? gcd(b, a % b) : a;}
int tr1[N], tr2[N], cnt;
int h[N], e[N], ne[N], w[N], idx;
int n, m, s;
ll d[N];
bool st[N];

void add(int x, int y, int z) {
	ne[idx] = h[x], e[idx] = y, w[idx] = z, h[x] = idx++;
}

int build1(int u, int l, int r) {
	tr1[u] = ++cnt;
	if (l == r) {
		return tr1[u] = l;
	}
	int mid = (l + r) >> 1;
	int lon = build1(u << 1, l, mid);
	int ron = build1(u << 1 | 1, mid + 1, r);
	add(tr1[u], lon, 0);
	add(tr1[u], ron, 0);
	return tr1[u];
}

int build2(int u, int l, int r) {
	tr2[u] = ++cnt;
	if (l == r) {
		return tr2[u] = l;
	}
	int mid = (l + r) >> 1;
	int lon = build2(u << 1, l, mid);
	int ron = build2(u << 1 | 1, mid + 1, r);
	add(lon, tr2[u], 0);
	add(ron, tr2[u], 0);
	return tr2[u];
}

void add1(int u, int l, int r, int L, int R, int x, int val) {
	if (l >= L && r <= R) {
		add(x, tr1[u], val);
		return;
	}
	int mid = (l + r) >> 1;
	if (L <= mid) add1(u << 1, l, mid, L, R, x, val);
	if (R > mid) add1(u << 1 | 1, mid + 1, r, L, R, x, val);
}

void add2(int u, int l, int r, int L, int R, int x, int val) {
	if (l >= L && r <= R) {
		add(tr2[u], x, val);
		return;
	}
	int mid = (l + r) >> 1;
	if (L <= mid) add2(u << 1, l, mid, L, R, x, val);
	if (R > mid) add2(u << 1 | 1, mid + 1, r, L, R, x, val);
}

void dij() {
	memset(d,0x3f,sizeof d);
	d[s] = 0;
	priority_queue<PII, vector<PII>, greater<PII> > Q;
	Q.push(PII(0, s));
	while (Q.size()) {
		PII p = Q.top();
		Q.pop();
		int t = p.second;
		if (st[t]) continue;
		st[t] = true;
		for (int i = h[t]; i != -1; i = ne[i]) {
			int j = e[i];
			if (d[j] > d[t] + w[i]) {
				d[j] = d[t] + w[i];
				Q.push(PII(d[j], j));
			}
		}
	}
}

int main() {
	memset(h, -1, sizeof h);
	read(n), read(m), read(s);
	cnt = n;
	build1(1, 1, n);
	build2(1, 1, n);
	while (m--) {
		int op, x, l, r, w;
		read(op), read(x), read(l), read(r);
		if (op == 1) add(x, l, r);
		else {
			read(w);
			if (op == 2) add1(1, 1, n, l, r, x, w);
			else add2(1, 1, n, l, r, x, w);
		}
	}
	dij();
	for (int i = 1; i <= n; i++) {
		if (d[i] == INF) d[i] = -1;
		printf("%lld ", d[i]);
	}
	return 0;
}

标签：787D,return,int,线段,CodeForces,mid,read,建图,const
来源： https://blog.csdn.net/tlyzxc/article/details/119461012