题解【LOJ3087】「GXOI / GZOI2019」旅行者

题面

对正图和反图分别跑一次最短路，求出离每个点最近的关键点。

对于每一条边，如果离两个端点最近的点不同，那么与答案取 \(\min\) 即可。

类似的题目有 牛客 IOI 周赛19 - 提高组 A. 基站。

代码可以参考一下，毕竟这也是一类经典套路。

#include 
#define DEBUG fprintf(stderr, "Passing [%s] line %d\n", __FUNCTION__, __LINE__)
#define File(x) freopen(x".in","r",stdin); freopen(x".out","w",stdout)

using namespace std;

typedef long long LL;
typedef pair  PII;
typedef pair  PLI;
typedef pair  PIII;

template 
inline T gi()
{
	T f = 1, x = 0; char c = getchar();
	while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
	while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
	return f * x;
}

const int INF = 0x3f3f3f3f, N = 100003, M = 1000003;

int T, n, m, k;
int tot, head[N], headc[N], ver[M], nxt[M], edge[M];
int g[2][N];
LL dis[2][N];
bool vis[N], isk[N];
LL ans;

inline void add(int h[], int u, int v, int w) {ver[++tot] = v, edge[tot] = w, nxt[tot] = h[u], h[u] = tot;}

inline void Dij(bool fl)
{
	priority_queue  q;
	memset(dis[fl], 0x3f, sizeof dis[fl]); memset(vis, false, sizeof vis);
	for (int i = 1; i <= n; i+=1) if (isk[i]) q.push({0, i}), dis[fl][i] = 0, g[fl][i] = i;
	while (!q.empty())
	{
		int u = q.top().second; q.pop();
		if (vis[u]) continue; vis[u] = true;
		for (int i = fl ? headc[u] : head[u]; i; i = nxt[i])
		{
			int v = ver[i], w = edge[i];
			if (dis[fl][v] > dis[fl][u] + w)
			{
				dis[fl][v] = dis[fl][u] + w;
				g[fl][v] = g[fl][u];
				q.push({-dis[fl][v], v});
			}
		}
	}
}

int main()
{
//	File("");
	T = gi  ();
	while (T--)
	{
		n = gi  (), m = gi  (), k = gi  ();
		memset(head, 0, sizeof head); memset(headc, 0, sizeof headc); tot = 0;
		for (int i = 1; i <= m; i+=1)
		{
			int u = gi  (), v = gi  (), w = gi  ();
			add(head, u, v, w), add(headc, v, u, w);
		}
		for (int i = 1; i <= n; i+=1) isk[i] = false;
		for (int i = 1; i <= k; i+=1) isk[gi  ()] = true;
		ans = 1e18;
		Dij(0); Dij(1);
		for (int u = 1; u <= n; u+=1)
			for (int i = head[u]; i; i = nxt[i])
			{
				int v = ver[i];
				if (g[0][u] != g[1][v]) ans = min(ans, edge[i] + dis[0][u] + dis[1][v]);
			}
		printf("%lld\n", ans);
	}
	return 0;
}