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\n", ans);}
}