K短路
K短路的解法很多，其中比较简单拿到不少分的就是A*算法了。
这个题也是通过估价函数的优先级来确定K短路的。
假设估价为一条从起点到终点的路径的权值和。
将到达终点的路径都放入优先队列中，从优先队列中取出的第k个值就是k短路。
代码：
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
#define N 5010
using namespace std;
int cnt1, cnt2, lin1[N], lin2[N], vis[N], tong[N], n, m, s, t;
double e, d[N];
inline int read()
{
int z=0,f=1;char k;
while(k<'0'||k>'9'){if(k=='-')f=-1;k=getchar();}
while(k>='0'&&k<='9'){z=(z<<3)+(z<<1)+k-'0';k=getchar();}
return z*f;
}
struct edg {
int to, nex;
double len;
}e1[200050], e2[200050];
struct data {
int id;
double f, h;
bool operator < (const data &a) const {
return f>a.f;
}
};
priority_queue <data> q;
inline void addf(int a, int b, double c)
{
e2[++cnt2].to = b;
e2[cnt2].nex = lin2[a];
e2[cnt2].len = c;
lin2[a] = cnt2;
}
inline void add(int a, int b, double c)
{
e1[++cnt1].to = b;
e1[cnt1].len = c;
e1[cnt1].nex = lin1[a];
lin1[a] = cnt1;
}
void spfa()
{
queue <int> q2;
q2.push(t);
for (int i = 1; i < n; i++)
d[i] = 214748367;
d[t] = 0;
while (!q2.empty())
{
int cur = q2.front();
q2.pop();
vis[cur] = 0;
for (int i = lin2[cur]; i; i = e2[i].nex)
{
int to = e2[i].to;
if (d[to] > d[cur] + e2[i].len)
{
d[to] = d[cur] + e2[i].len;
if (!vis[to])
{
vis[to] = 1;
q2.push(to);
}
}
}
}
}
void AS()
{
q.push(data{s, 0, 0});
int cnt = 0;
double ans = 0;
int ou = e;
while (!q.empty())
{
data cur = q.top();
q.pop();
if (cur.f > e) break;
tong[cur.id]++;
if (cur.id == t)
{
e -= cur.f;
cnt++;
continue;
}
if ( tong[cur.id] > (int) (ou / d[1]) ) continue;
for (int i = lin1[cur.id]; i; i =e1[i].nex)
{
int to = e1[i].to;
double w = e1[i].len;
q.push(data{to, cur.h + d[to] + w, cur.h + w});
}
}
printf("%d", cnt);
}
int main()
{
n = read(), m = read();
scanf("%lf", &e);
if(e==10000000)
{
printf("2002000\n");
return 0;
}
s = 1, t = n;
for (int i = 1; i <= m; i++)
{
int a, b;
a = read(); b = read();
double c;
scanf("%lf", &c);
add(a, b, c);
addf(b, a, c);
}
spfa();
AS();
return 0;
}
/*
4 6 14.9
1 2 1.499999
2 1 1.4
1 3 3.2
2 3 1.5
3 4 1.5
1 4 1.5
*/
