【图论经典题目讲解】洛谷 P2149 Elaxia的路线

P2149 Elaxia的路线

D e s c r i p t i o n \mathrm{Description} Description

给定 n n n 个点， m m m 条边的无向图，求 2 2 2 个点对间最短路的最长公共路径

S o l u t i o n \mathrm{Solution} Solution

最短路有可能不唯一，所以公共路径的长度就有可能不同。

将 2 2 2 条最短路都会经过的边（包括同向和异向）记录出来，并建立 1 1 1 个新图，那么由于最短路（可以看做一条链）一定不会走环，故新图必定是一个 有向无环图 （简称 D A G \mathrm{DAG} DAG），而 D A G \mathrm{DAG} DAG 图上就可以跑 DP 来求解最长链，由于找出的是 2 2 2 条最短路都经过的边，所以最长链其实就是 2 2 2 条最短路的最长公共路径。

故，该问题得以解决。

C o d e Code Code

#include <bits/stdc++.h>
#define int long long

using namespace std;

typedef pair<int, int> PII;
typedef long long LL;

const int SIZE = 1e6 + 10;

int N, M;
int X1, Y1, X2, Y2;
int h[SIZE], hs[SIZE], e[SIZE], ne[SIZE], w[SIZE], idx;
int D[4][SIZE], Vis[SIZE], in[SIZE], q[SIZE], hh, tt = -1;
int F[SIZE];

void add(int h[], int a, int b, int c)
{
	e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx ++;
}

void Dijkstra(int Start, int dist[])
{
	for (int i = 1; i <= N; i ++)
		dist[i] = 1e18, Vis[i] = 0;
	priority_queue<PII, vector<PII>, greater<PII>> Heap;
	Heap.push({0, Start}), dist[Start] = 0;

	while (Heap.size())
	{
		auto Tmp = Heap.top();
		Heap.pop();

		int u = Tmp.second;
		if (Vis[u]) continue;
		Vis[u] = 1;

		for (int i = h[u]; ~i; i = ne[i])
		{
			int j = e[i];
			if (dist[j] > dist[u] + w[i])
			{
				dist[j] = dist[u] + w[i];
				Heap.push({dist[j], j});
			}
		}
	}
}

void Topsort()
{
	hh = 0, tt = -1;
	for (int i = 1; i <= N; i ++)
		if (!in[i])
			q[ ++ tt] = i;
	while (hh <= tt)
	{
		int u = q[hh ++];

		for (int i = hs[u]; ~i; i = ne[i])
		{
			int j = e[i];
			if ((-- in[j]) == 0)
				q[ ++ tt] = j;
		}
	}
}

signed main()
{
	cin.tie(0);
	cout.tie(0);
	ios::sync_with_stdio(0);

	memset(h, -1, sizeof h);
	memset(hs, -1, sizeof hs);

	cin >> N >> M >> X1 >> Y1 >> X2 >> Y2;

	int u, v, c;
	while (M --)
	{
		cin >> u >> v >> c;
		add(h, u, v, c), add(h, v, u, c);
	}

	Dijkstra(X1, D[0]), Dijkstra(Y1, D[1]);
	Dijkstra(X2, D[2]), Dijkstra(Y2, D[3]);

	for (int i = 1; i <= N; i ++)
		for (int j = h[i]; ~j; j = ne[j])
		{
			int k = e[j];
			if (D[0][i] + w[j] + D[1][k] == D[0][Y1] && D[2][i] + w[j] + D[3][k] == D[2][Y2])
				add(hs, i, k, w[j]), in[k] ++;
		}

	Topsort();

	for (int it = 0; it <= tt; it ++)
	{
		int i = q[it];
		for (int j = hs[i]; ~j; j = ne[j])
		{
			int k = e[j];
			F[k] = max(F[k], F[i] + w[j]);
		}
	}

	int Result = 0;
	for (int i = 1; i <= N; i ++)
		Result = max(Result, F[i]);

	memset(hs, -1, sizeof hs);
	memset(F, 0, sizeof F);
	memset(in, 0, sizeof in);

	for (int i = 1; i <= N; i ++)
		for (int j = h[i]; ~j; j = ne[j])
		{
			int k = e[j];
			if (D[0][i] + w[j] + D[1][k] == D[0][Y1] && D[3][i] + w[j] + D[2][k] == D[2][Y2])
				add(hs, i, k, w[j]), in[k] ++;//, cout << i << " " << k << endl;
		}

	Topsort();

	for (int it = 0; it <= tt; it ++)
	{
		int i = q[it];
		for (int j = hs[i]; ~j; j = ne[j])
		{
			int k = e[j];
			F[k] = max(F[k], F[i] + w[j]);
		}
	}

	for (int i = 1; i <= N; i ++)
		Result = max(Result, F[i]);

	cout << Result << endl;

	return 0;
}