CF786B Legacy
给定一张有向图,有三种边:
- \(u\to v\) ,边权为 \(w\)
- \(u\to [l,\ r]\) ,边权为 \(w\)
- \([l,\ r]\to u\) ,边权为 \(w\)
\(n,\ m\leq10^5,\ w\leq10^9\)
线段树优化建图,最短路
线段树优化建图板子题,注意空间大小
由于有 \(n\log n\) 条边, 且普通 dijkstra 时间复杂度为 \(O(m\log n)\)
所以时间复杂度 \(O(n\log^2 n)\)
代码
#include <bits/stdc++.h>
using namespace std;
#define mid ((l + r) >> 1)
#define lson ls[k], l, mid
#define rson rs[k], mid + 1, r
typedef long long ll;
typedef pair <ll, int> pii;
const int maxn = 1e5 + 10, maxm = maxn << 3;
ll dis[maxm];
int n, m, s, tot, rt_i, rt_o, h[maxm], ls[maxm], rs[maxm];
struct edges {
int nxt, to, w;
edges() {}
edges(int x, int y, int z) :
nxt(x), to(y), w(z) {}
} e[maxn * 30];
void addline(int u, int v, int w) {
static int cnt;
e[++cnt] = edges(h[u], v, w), h[u] = cnt;
}
void build_i(int &k, int l, int r) {
k = ++tot;
if (l == r) {
addline(k, l, 0); return;
}
build_i(lson), build_i(rson);
addline(k, ls[k], 0), addline(k, rs[k], 0);
}
void build_o(int &k, int l, int r) {
k = ++tot;
if (l == r) {
addline(l, k, 0); return;
}
build_o(lson), build_o(rson);
addline(ls[k], k, 0), addline(rs[k], k, 0);
}
void ins_i(int k, int l, int r, int ql, int qr, int u, int x) {
if (ql <= l && r <= qr) {
addline(u, k, x); return;
}
if (ql <= mid) ins_i(lson, ql, qr, u, x);
if (qr > mid) ins_i(rson, ql, qr, u, x);
}
void ins_o(int k, int l, int r, int ql, int qr, int u, int x) {
if (ql <= l && r <= qr) {
addline(k, u, x); return;
}
if (ql <= mid) ins_o(lson, ql, qr, u, x);
if (qr > mid) ins_o(rson, ql, qr, u, x);
}
int main() {
scanf("%d %d %d", &n, &m, &s);
tot = n;
build_i(rt_i, 1, n);
build_o(rt_o, 1, n);
for (int i = 1; i <= m; i++) {
int op, u, l, r, w;
scanf("%d %d %d %d", &op, &u, &l, &r);
if (op == 1) {
addline(u, l, r);
} else if (op == 2) {
scanf("%d", &w);
ins_i(rt_i, 1, n, l, r, u, w);
} else {
scanf("%d", &w);
ins_o(rt_o, 1, n, l, r, u, w);
}
}
static priority_queue <pii, vector <pii>, greater <pii> > Q;
memset(dis, 0x3f, sizeof dis);
dis[s] = 0, Q.push(pii(0, s));
while (!Q.empty()) {
pii p = Q.top();
int u = p.second;
Q.pop();
if (dis[u] < p.first) continue;
for (int i = h[u]; i; i = e[i].nxt) {
int v = e[i].to;
if (dis[v] > dis[u] + e[i].w) {
dis[v] = dis[u] + e[i].w;
Q.push(pii(dis[v], v));
}
}
}
for (int i = 1; i <= n; i++) {
printf("%I64d ", dis[i] < 1ll << 60 ? dis[i] : -1);
}
return 0;
}