제출 #636897

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
636897		chjiao	Commuter Pass (JOI18_commuter_pass)	C++17	100 / 100	506 ms	31588 KiB

commuter_pass

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;

vector<pair<ll, ll>> graph[100001];
ll du[100001], dv[100001], ds[100001], dp[2][100001], ans;
bool visited[100001];

void dijkstra1(ll start, ll arr[]) {
	fill(visited, visited + 100001, false);  //初始化visited为false

	priority_queue<pair<ll, ll>> pq;
	pq.push({0, start});
	while (!pq.empty()) {
		ll c, node;
		tie(c, node) = pq.top();  //c: 从 start 到 node 的cost
		pq.pop();

		if (!visited[node]) {
			arr[node] = -c;   //距离
			visited[node] = true;
			for (auto& i : graph[node]) pq.push({c - i.second, i.first});
		}
	}
}

void dijkstra2(ll start, ll end) {
	fill(dp[0], dp[0] + 100001, LLONG_MAX / 2);  //初始化dp[0], minimum distance from u to any node x such that x is in a path from s to i in the DAG
	fill(dp[1], dp[1] + 100001, LLONG_MAX / 2);  //初始化dp[1], minimum distance from v to any node x such that x is in a path from s to i in the DAG
	fill(visited, visited + 100001, false);  //初始化visited为false

	priority_queue<pair<ll, pair<ll, ll>>> pq;
	pq.push({0, {start, 0}});   //cost, curr node, pre node
	dp[0][0] = dp[1][0] = LLONG_MAX/ 2;
	while (!pq.empty()) {
		ll c, node, par;
		pair<ll, ll> p;
		tie(c, p) = pq.top();
		tie(node, par) = p;
		pq.pop();

		if (!visited[node]) {
			visited[node] = true;
			ds[node] = -c;
			dp[0][node] = min(du[node], dp[0][par]);
			dp[1][node] = min(dv[node], dp[1][par]);
			for (auto i : graph[node]) pq.push({c - i.second, {i.first, node}});  //
		} else if (-c == ds[node]) { //另外一条最短路近
			if (min(du[node], dp[0][par]) + min(dv[node], dp[1][par]) <= dp[0][node] + dp[1][node]) {
				dp[0][node] = min(du[node], dp[0][par]);
				dp[1][node] = min(dv[node], dp[1][par]);
			}
		}
	}

	ans = min(ans, dp[0][end] + dp[1][end]);
}

int main() {
	iostream::sync_with_stdio(false);
	cin.tie(0);
    // freopen("CommuterPass.in", "r", stdin);
    // freopen("CommuterPass.out", "w", stdout); 
	ll n, m, s, t, u, v;
	cin >> n >> m >> s >> t >> u >> v;
	for (int i = 0; i < m; i++) {
		ll a, b, c;
		cin >> a >> b >> c;
		graph[a].push_back({b, c});
		graph[b].push_back({a, c});
	}

	dijkstra1(u, du);  //u出发，到所有其他点的距离，保存在du中。
	dijkstra1(v, dv);  //v出发，到所有其他点的距离，保存在dv中。

	ans = du[v];   //ans是u到v的最短路径

	dijkstra2(s, t);
	dijkstra2(t, s);

	cout << ans << '\n';
	return 0;
}

• Subtask 1: 16 / 16
• Subtask 2: 15 / 15
• Subtask 3: 24 / 24
• Subtask 4: 45 / 45