https://www.luogu.org/problemnew/show/P4568

题目中 k 的大小只有 10，我们可以考虑建立分层图跑最短路

相同层中 a -> b 的权值仍为 val，不同层中 a -> b 的权值为 0，相当于免费乘坐了一次飞机

写一个最短路就可以啦

#include <bits/stdc++.h>
#define CIOS ios::sync_with_stdio(false);
#define For(i, a, b) for(register int i = a; i <= b; i++)
#define Forr(i, a, b) for(register int i = a; i >= b; i--)
using namespace std;typedef unsigned long long ull;
typedef long long ll;template <typename _T>
inline void read(_T &f) {f = 0; _T fu = 1; char c = getchar();while(c < '0' || c > '9') { if(c == '-') fu = -1; c = getchar(); }while(c >= '0' && c <= '9') { f = (f << 3) + (f << 1) + (c & 15); c = getchar(); }f *= fu;
}template <typename T>
void print(T x) {if(x < 0) putchar('-'), x = -x;if(x < 10) putchar(x + 48);else print(x / 10), putchar(x % 10 + 48);
}template <typename T>
void print(T x, char t) {print(x); putchar(t);
}const int N = 4e5 + 5, M = 4e6 + 5;struct Edge {int u, v, next, val;
}G[M];int head[N];
int n, m, k, s, t, tot;inline void addedge(int u, int v, int val) {G[++tot] = (Edge) {u, v, head[u], val}, head[u] = tot;
}int dis[N];
priority_queue < pair <int, int> > Q;
void dij(int s) {memset(dis, 0x3f, sizeof(dis));dis[s] = 0; Q.push(make_pair(0, s));while(!Q.empty()) {pair <int, int> t = Q.top(); Q.pop();if(-t.first > dis[t.second]) continue;int u = t.second;for(register int i = head[u]; i; i = G[i].next) {int v = G[i].v;if(dis[v] > dis[u] + G[i].val) {dis[v] = dis[u] + G[i].val;Q.push(make_pair(-dis[v], v));}}}
}int main() {read(n); read(m); read(k);read(s); read(t);while(m--) {int a, b, c;read(a); read(b); read(c);for(register int i = 1; i <= k + 1; i++) addedge((i - 1) * n + a, (i - 1) * n + b, c), addedge((i - 1) * n + b, (i - 1) * n + a, c);for(register int i = 1; i <= k; i++) addedge((i - 1) * n + a, i * n + b, 0), addedge((i - 1) * n + b, i * n + a, 0);}dij(s); int ans = 0x7fffffff;for(register int i = 1; i <= k + 1; i++) ans = min(ans, dis[(i - 1) * n + t]);cout << ans << endl;return 0;
}