2019 牛客多校 第四场 J、free 分层图最短路

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 
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 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;
}