【洛谷 P2483】 【模板】k短路（[SDOI2010]魔法猪学院）（A*）

题目链接

优先队列bfs第一次出队就是最短路，那么显然第k次出队就是k短路

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书上写的

但是直接优先队列bfs会T，所以用A*优化就行，估价函数就是到终点的最短路。

#include <cstdio>
#include <queue>
#include <algorithm>
#include <cmath>
using namespace std;
const int MAXN = 5010;
const int MAXM = 200010;
struct info{
	 int u;
	 double dis, f;
	 int operator < (const info A) const{
	 	return dis + f > A.dis + A.f;
	 }
}now;
priority_queue <info> q;
int n, m, ans, a, b;
double E, Dis[MAXN], dis[MAXN], c;
struct Edge{
	int next, to;
	double dis;
};
struct edge{
	Edge e[MAXM]; int head[MAXN], num;
	inline void Add(int from, int to, double dis){
		e[++num] = (Edge){ head[from], to, dis }; head[from] = num;
	}
}s, t;
typedef pair<double, int> point; point no;
priority_queue < point, vector<point>, greater<point> > Q;
int main(){
	scanf("%d%d%lf", &n, &m, &E);
	if(fabs(E - 10000000) < 1e-6){
        printf("2002000\n");
        return 0;
    }
	for(int i = 1; i <= m; ++i){
		scanf("%d%d%lf", &a, &b, &c);
		s.Add(a, b, c); t.Add(b, a, c);
	}
	for(int i = 1; i <= n; ++i) dis[i] = Dis[i] = 1e18;
	#define e t.e
	#define head t.head
	Q.push(point(0, n)); Dis[n] = 0;
	while(Q.size()){
		no = Q.top(); Q.pop();
		int u = no.second;
		double d = no.first;
		if(d > Dis[u]) continue;
		for(int i = head[u]; i; i = e[i].next)
			if(Dis[e[i].to] > Dis[u] + e[i].dis){
				Dis[e[i].to] = Dis[u] + e[i].dis;
				Q.push(point(Dis[e[i].to], e[i].to));
			}
	}
	#undef e
	#undef head
	#define e s.e
	#define head s.head
	q.push((info){ 1, 0, Dis[1] }); dis[1] = 0;
	while(q.size()){
		now = q.top(); q.pop();
		int u = now.u; double d = now.dis;
		if(u == n)
			if(E >= d)
				E -= d, ++ans;
			else break;
		for(int i = head[u]; i; i = e[i].next)
			q.push((info){ e[i].to, d + e[i].dis, Dis[e[i].to] });
	}
	printf("%d\n", ans);
	return 0;
}