图论-第k短路

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
", Ans);
	return 0;
}