【分层最短路】Magical Girl Haze

https://nanti.jisuanke.com/t/31001

有K次机会可以让一条边的权值变为0，求最短路。

在存储单源最短路的数组上多开一维状态，d[i][k]表示走到序号i的点，且让k条边权值为0时的最短路。
对于每个待更新的点，尝试不置零此边的状态和置零此边的状态，分别压入优先队列去更新其他状态。

另外，此题由于有重边，需要先去重，保留同起始点最短的边。

代码：

#include <iostream>
#include <algorithm>
#include <queue>
#include <cstring>
#include <vector>
#include <cstdio>
using namespace std;
typedef long long ll;
const ll INF = 0x3f3f3f3f3f3f3f3fLL;
const int MAX_V = 100005;
const int MAX_E = 200005;
int N, M, K;
struct Dijkstra {
	struct Edge {
		int to, cost, next;
	} es[MAX_E];
	struct Node {
		int u, k;
		ll d;
		Node(int u, ll d, int k) : u(u), d(d), k(k) {}
		bool operator< (const Node& n) const {
			return d > n.d;
		}
	};
	int head[MAX_V];
	int V, E;
	ll d[MAX_V][15];
	bool vis[MAX_V][15];
	void init(int V) {
		this->V = V;
		this->E = 0;
		memset(head, -1, sizeof head);
	}
	void addEdge(int u, int v, int w) {
		es[E].to = v;
		es[E].cost = w;
		es[E].next = head[u];
		head[u] = E++;
	}
	void dijkstra(int s) {
		priority_queue <Node> Q;
		memset(d, 0x3f, sizeof d);
		memset(vis, 0, sizeof(vis));
		d[s][0] = 0;
		Q.push(Node(s, 0, 0));
		while (!Q.empty()) {
			int u = Q.top().u, k = Q.top().k;
			Q.pop();
			if (vis[u][k])
				continue;
			vis[u][k] = true;
			for (int i = head[u]; i != -1; i = es[i].next) {
				int v = es[i].to, w = es[i].cost;
				if (d[v][k] > d[u][k] + w) {
					d[v][k] = d[u][k] + w;
					Q.push(Node(v, d[v][k], k));
				}
				if (k + 1 <= K) {
					if (d[v][k + 1] > d[u][k]) {
						d[v][k + 1] = d[u][k];
						Q.push(Node(v, d[v][k + 1], k + 1));
					}
				}
			}
		}
	}
} dijk;
struct Elem {
	int u, v, w;
	bool operator< (const Elem& e) const {
		if (u == e.u && v == e.v) {
			return w < e.w;
		}
		if (u == e.u) {
			return v < e.v;
		}
		return u < e.u;
	}
} e[MAX_E];
int main() {
	int T;
	scanf("%d", &T);
	while (T--) {
		scanf("%d%d%d", &N, &M, &K);
		dijk.init(N);
		for (int i = 0; i < M; i++) {
			scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].w);
		}
		sort(e, e + M);
		int preu = 0, prev = 0;
		for (int i = 0; i < M; i++) {
			if (preu != e[i].u || prev != e[i].v) {
				dijk.addEdge(e[i].u, e[i].v, e[i].w);
				preu = e[i].u, prev = e[i].v;
			}
		}
		dijk.dijkstra(1);
		ll ans = INF;
		for (int i = 0; i <= K; i++) {
			ans = min(ans, dijk.d[N][i]);
		}
		printf("%lld
", ans);
	}
}