「AMPPZ2014」Petrol

传送门：
这是一道bzoj权限题
Luogu团队题链接

解题思路

首先对于每一个点 (x) 预处理出 (nr[x]) 和 (dis[x])，分别表示离 (x) 最近的加油站以及该段距离。
这个过程可以用多源 ( ext{Dijkstra}) 处理。
然后对于每一条原图中的边 ((u, v, w))（(nr[u] e nr[v])）
改为这样一条边：((nr[u], nr[v], dis[u] + dis[v] + w))
然后只要离线用并查集维护一下连通性即可。

细节注意事项

• 最短路不要写挂啊

参考代码

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cctype>
#include <cmath>
#include <ctime>
#include <queue>
#define rg register
#define pii pair < int, int >
using namespace std;
template < typename T > inline void read(T& s) {
	s = 0; int f = 0; char c = getchar();
	while (!isdigit(c)) f |= (c == '-'), c = getchar();
	while (isdigit(c)) s = s * 10 + (c ^ 48), c = getchar();
	s = f ? -s : s;
}

const int _ = 200010;
const int __ = 400010;

int tot, head[_], nxt[__], ver[__], w[__];
inline void Add_edge(int u, int v, int d)
{ nxt[++tot] = head[u], head[u] = tot, ver[tot] = v, w[tot] = d; }

int q, n, m, k, c[_], vis[_], dis[_], nr[_];
struct node{ int u, v, d, id; }ask[_], g[__], e[__]; int ans[_];

inline bool cmp(const node& x, const node& y) { return x.d < y.d; }

inline void Dijkstra() {
	static priority_queue < pii > Q;
	memset(dis, 0x3f, sizeof dis);
	for (rg int i = 1; i <= k; ++i)
		Q.push(make_pair(dis[c[i]] = 0, c[i])), nr[c[i]] = c[i];
	while (!Q.empty()) {
		int u = Q.top().second; Q.pop();
		if (vis[u]) continue; vis[u] = 1;
		for (rg int i = head[u]; i; i = nxt[i]) {
			int v = ver[i];
			if (dis[v] > dis[u] + w[i])
				dis[v] = dis[u] + w[i], nr[v] = nr[u], Q.push(make_pair(-dis[v], v));
		}
	}
}

int fa[_];

inline int findd(int x) { return fa[x] == x ? x : fa[x] = findd(fa[x]); }

inline void merge(int x, int y) { fa[findd(x)] = findd(y); }

int main() {
#ifndef ONLINE_JUDGE
	freopen("in.in", "r", stdin);
#endif
	read(n), read(k), read(m);
	for (rg int i = 1; i <= k; ++i) read(c[i]);
	for (rg int u, v, d, i = 1; i <= m; ++i)
		read(u), read(v), read(d), Add_edge(u, v, d), Add_edge(v, u, d), g[i] = (node) { u, v, d };
	Dijkstra();
	int cnt = 0;
	for (rg int u, v, i = 1; i <= m; ++i) {
		u = g[i].u, v = g[i].v;
		if (nr[u] != nr[v]) e[++cnt] = (node) { nr[u], nr[v], dis[u] + dis[v] + g[i].d };
	}
	read(q);
	for (rg int i = 1; i <= q; ++i)
		read(ask[i].u), read(ask[i].v), read(ask[i].d), ask[i].id = i;
	sort(ask + 1, ask + q + 1, cmp);
	sort(e + 1, e + cnt + 1, cmp);
	for (rg int i = 1; i <= n; ++i) fa[i] = i;
	for (rg int i = 1, j = 1; i <= q; ++i) {
		while (j <= cnt && e[j].d <= ask[i].d) merge(e[j].u, e[j].v), ++j;
		ans[ask[i].id] = findd(ask[i].u) == findd(ask[i].v);
	}
	for (rg int i = 1; i <= q; ++i) puts(ans[i] ? "TAK" : "NIE");
	return 0;
}

完结撒花 (qwq)