【图论经典题目讲解】洛谷 P2149 Elaxia的路线

本文主要是介绍【图论经典题目讲解】洛谷 P2149 Elaxia的路线，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

P2149 Elaxia的路线

$\mathrm{Description}$

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

$\mathrm{Solution}$

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 $2$ 条最短路都会经过的边（包括同向和异向）记录出来，并建立 $1$ 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 $\mathrm{DAG}$ ），而 $\mathrm{DAG}$ 图上就可以跑 DP 来求解最长链，由于找出的是 $2$ 条最短路都经过的边，所以最长链其实就是 $2$ 条最短路的最长公共路径。

故，该问题得以解决。

$C o d e$

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