基本模型:在图上,有k次机会可以直接通过一条边而不计算边权,问起点与终点之间的最短路径。
#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 1e5+10;
const ll INF = 1e17;
const int M = 2e5+10;
ll cnt, k, n, m, s, t, v, u, ww;//起点s到终点t有k次免费走的机会
ll dist[N][30], w[M];
int fir[N], nex[M], to[M];
struct Nod {
ll x, k, d;
bool operator<(const Nod &a)const {
return d > a.d;
}
};
void add(ll u, ll v, ll ww) {
nex[++ cnt] = fir[u];
fir[u] = cnt;
to[cnt] = v;
w[cnt] = ww;
}
ll dijstra() {
memset(dist,127,sizeof(dist));
priority_queue<Nod> que;
que.push((Nod) {s,0,0});
dist[s][0] = 0;
while(que.size()) {
Nod e = que.top();
que.pop();
if(e.d != dist[e.x][e.k]) continue;
if(e.x == t) return e.d;
for(int i = fir[e.x]; i; i = nex[i]) {
if(dist[to[i]][e.k] > dist[e.x][e.k] + w[i]) {
dist[to[i]][e.k] = dist[e.x][e.k] + w[i];
que.push((Nod) {to[i], e.k, dist[to[i]][e.k]});
}
if(e.k < k && dist[to[i]][e.k+1] > dist[e.x][e.k]) {
dist[to[i]][e.k+1] = dist[e.x][e.k];
que.push((Nod) {to[i],e.k+1,dist[to[i]][e.k+1]});
}
}
}
}
void init() {
cnt = 0;
memset(nex,0,sizeof(nex));
memset(fir,0,sizeof(fir));
memset(to,0,sizeof(to));
}
int main() {
init();
scanf("%lld %lld %lld %lld %lld", &n, &m, &s, &t, &k);
for(int i = 1; i <= m; i ++) {
scanf("%lld%lld%lld", &v, &u, &ww);
if(u==v)continue;
add(v,u,ww);
add(u,v,ww);
}
printf("%lld\n",dijstra());
return 0;
}