A* 做法
\(f(p)=g(p)+h(p)\) ， \(f(p)\) 作为优先队列比较函数用来比较的值， \(g(p)\) 是当前路径到 \(p\) 的距离， \(h(p)\) 是 \(p\) 点到终点最短路（预处理可以得到）。
每个点出队次数 \(k\)，就说明当前找到的是到这个点的 \(k\) 短路。
关键代码
void astar(int bg)
{int cnt = 0;A.push(ast(bg, 0));while (!A.empty()) {ast p = A.top();A.pop();if (p.v == N) {if (p.w > E)break;E -= p.w, ++Ans;}for (int i = G[p.v].size() - 1; i >= 0; --i) {edge e = G[p.v][i];A.push(ast(e.v, p.w + e.w));}}return;
}一道裸题： 【SDOI2010】魔法猪学院
#include <cstdio>
#include <vector>
#include <algorithm>
#include <queue>using namespace std;typedef double db;const int _N = 5100;
const db INF = 1e9;struct edge {int v;db w;edge(int v = 0, db w = 0): v(v), w(w) { }
};struct data {int v;db w;data(int v = 0, db w = 0): v(v), w(w) { }bool operator < (const data &tmp)const{return w > tmp.w;}
};db E, dis[_N];struct ast {int v;db w;ast(int v = 0, db w = 0): v(v), w(w) { }bool operator < (const ast &tmp)const{return w + dis[v] > tmp.w + dis[tmp.v];}
};priority_queue<data> Q;
priority_queue<ast> A;
vector<edge> G[_N], H[_N];
int Ans, N, M;void Gins(int a, int b, db c) { G[a].push_back(edge(b ,c)); return; }void Hins(int a, int b, db c) { H[a].push_back(edge(b, c)); return; }void dfs(int x, db cost)
{if (cost + dis[x] > E) return;if (x == N) {A.push(cost);return;}for (int i = G[x].size() - 1; i >= 0; --i) {edge e = G[x][i];dfs(e.v, cost + e.w);}return;
}void init(int bg)
{for (int i = 1; i <= N; ++i)dis[i] = INF;Q.push(data(bg, dis[bg] = 0));while (!Q.empty()) {data p = Q.top();Q.pop();if (dis[p.v] != p.w) continue;for (int i = H[p.v].size() - 1; i >= 0; --i) {edge e = H[p.v][i];if (dis[e.v] > p.w + e.w)dis[e.v] = p.w + e.w, Q.push(data(e.v, dis[e.v]));}}return;
}void astar(int bg)
{int cnt = 0;A.push(ast(bg, 0));while (!A.empty()) {ast p = A.top();A.pop();if (p.v == N) {if (p.w > E)break;E -= p.w, ++Ans;}for (int i = G[p.v].size() - 1; i >= 0; --i) {edge e = G[p.v][i];A.push(ast(e.v, p.w + e.w));}}return;
}int main()
{scanf("%d%d%lf", &N, &M, &E);for (int i = 1; i <= M; ++i) {int a, b;db c;scanf("%d%d%lf", &a, &b, &c);Gins(a, b, c), Hins(b, a, c);}init(N);astar(1);printf("%d\n", Ans);return 0;
}
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