[机房测试]普通快乐

给定一个 \(n\) 个点 \(m\) 条边的无向图和 \(k\) 个关键点，求关键点间两两距离的最小值。

正解:

点两两距离最小值的套路是新建一个源点 \(S\) 和一个汇点 \(T\)，都向关键点连边权为0的边，此时 \(S\) 到 \(T\) 的最短路就是答案？注意到我们从 \(S\) 走到一个点会直接走到 \(T\)，最短路是0.所以我们考虑二进制分组，枚举位数 \(i\), 第 \(i\) 位为 1 的连向 \(S\)，否则连向 \(T\)，这样求出的就是两个集合之间的最短距离。显然每对点都会被枚举到。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <ctime>

using namespace std;

const int N = 200005;

int fr[N << 1], t[N << 1], w[N << 1];

struct edge {
	int head, to, nxt, val;
} ed[N << 2];

int en = 0, dis[N], n, m, k, ans = 0x3f3f3f3f;
int b[N], top = 0, mp[N];
int vis[N];
bool c[N];

struct node {
	int x, d;
	inline node(int X, int D) : x(X), d(D) { }
	inline bool operator > (const node &b) const { return d > b.d; }
	inline bool operator < (const node &b) const { return d < b.d; }
};

priority_queue<node, vector<node>, greater<node> > q;

inline int min_(int a, int b) {
	return a < b ? a : b;
}

inline int read() {
	register int s = 0; register char ch = getchar();
	while (!isdigit(ch)) ch = getchar();
	while (isdigit(ch)) s = (s << 1) + (s << 3) + (ch & 15), ch = getchar();
	return s;
}

inline void addedge(int from, int to, int val) {
	ed[++en].to = to; ed[en].val = val; ed[en].nxt = ed[from].head; ed[from].head = en;
}

inline void dijkstra(int s) {
	for (int i = 0; i <= en || i <= n + 2; ++i) ed[i].head = ed[i].to = ed[i].nxt = ed[i].val = 0;
	en = 0;
	for (int i = 1; i <= m; ++i) {
		addedge(fr[i], t[i], w[i]); addedge(t[i], fr[i], w[i]);
	}
	int S = n + 1, T = n + 2;
	for (int i = 1; i <= k; ++i) {
		if ((i >> s) & 1) addedge(S, b[i], 0);
		else addedge(b[i], T, 0);
	}
	for (int i = 1; i <= n + 2; ++i) dis[i] = 0x3f3f3f3f, vis[i] = 0;
	dis[S] = 0; q.push(node(S, 0));
	while (!q.empty()) {
		int x = q.top().x; q.pop();
		if (vis[x] == 1) continue; vis[x] = 1;
		for (int i = ed[x].head; i; i = ed[i].nxt) {
			int v = ed[i].to;
			if (dis[v] > dis[x] + ed[i].val) {
				q.push(node(v, dis[v] = dis[x] + ed[i].val));
			}
		}
	} if (dis[T]) ans = min_(ans, dis[T]);
}

int main() {
//	freopen("path.in", "r", stdin); freopen("path.out", "w", stdout);
	n = read(); m = read(); k = read();
	for (int i = 1; i <= m; ++i) {
		fr[i] = read(); t[i] = read(); w[i] = read();
	} for (int i = 1; i <= k; ++i) {
		c[b[i] = read()] = 1; mp[b[i]] = i;
	}
	for (int i = 0; (1 << i) <= k; ++i) dijkstra(i);
	printf("%d", ans); return 0;
}