[KEYENCE Programming Contest 2023 Autumn（AtCoder Beginner Contest 325） E

KEYENCE Programming Contest 2023 Autumn（AtCoder Beginner Contest 325） - AtCoder E

E - Our clients, please wait a moment (atcoder.jp)(分层图最短路)

因为只能从坐公司汽车切换到做火车,所以我们可以考虑采用分层图最短路.

对于坐公司汽车和坐火车分别建立\(G_0\)和\(G_1\)两张图,且因为可以在任意时刻任意地点从公司汽车切换到火车,所以我们可以对于每个地点建立\(\forall x(G_0(x) \rightarrow G_1(x))\)的边权为\(0\)的有向边,\(G_0\)为第一层,有节点\(1 \sim n\),即公司汽车的路线;第二层有节点\(1+n \sim n + n\),即火车的路线,两层节点间\(i\)与\(i+n\)也用权值为0的有向边连起来,起点为第一层的\(1\),终点就是第二层的\(n+n\),注意因为分层,所以分层图\(dis\)范围初始化是\(K\)(\(K\)层)\(\times N + N\)(\(N\)个点).

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';

using namespace std;
using i64 = long long;

typedef pair<i64, i64> PII;

const int MAXV = 1e4 + 1;
struct edge {
	i64 to, cost;
};

vector<edge> G[MAXV];
i64 d[3010];

void dijkstra(int s) {

	priority_queue<PII, vector<PII>, greater<PII> > que;

	memset(d, 0x3f3f3f3f, sizeof d);
	d[s] = 0;
	que.push(PII(0, s));
	while (!que.empty())
	{
		PII p = que.top(); que.pop();
		int v = p.second;
		if (d[v] < p.first) continue;
		for (int i = 0; i < G[v].size(); i++) {
			edge e = G[v][i];
			if (d[e.to] > d[v] + e.cost) {
				d[e.to] = d[v] + e.cost;
				que.push(PII(d[e.to], e.to));
			}
		}
	}
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);

	int A, B, C, N;
	cin >> N >> A >> B >> C;
	vector D(N + 1, vector<i64>(N + 1));
	for (int i = 1; i <= N; i ++)
		for (int j = 1; j <= N; j ++)
			cin >> D[i][j];

	for (int i = 1; i <= N; i ++) {
		for (int j = 1; j <= N; j ++) {
			if (i == j) continue;
			G[i].push_back(edge{j, D[i][j] * A});
			G[i].push_back(edge{i + N, 0});
			G[i + N].push_back(edge{j + N, D[i][j] * B + C});
		}
	}

	dijkstra(1);

	cout << d[N + N] << '\n';

	return 0;
}

参考资料

浅析分层图最短路 - feather02的博客 - 洛谷博客 (luogu.com.cn)

ABC325E 题解 - Welcome to CultReborn's Blog - 洛谷博客 (luogu.com.cn)

分层图复习 / abc325e - 可可爱爱 - 洛谷博客 (luogu.com.cn)