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
");
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
", ans);
return 0;
}