ACM-ICPC 2018 沈阳赛区网络预赛 D. Made In Heaven A* K短路

题解

题目大意，给一个有向图，问起点到终点的k短路代价是否小于t。

使用A*算法求解，由于是有向图使用dijkstra得到其它所有点到终点的最短路时需要建立反向边。
给定的有向图不一定能从起点到达终点，在使用A*之前先判断是否可达，否则会造成死循环。

AC代码

#include <stdio.h>
#include <bits/stdc++.h>
#define fst first
#define sed second
using namespace std;
typedef long long ll;

const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
const int N = 1e4 + 10;
int n, m, st, ed, k, t;
bool vis[N];
ll dis[N];

struct edge
{
	int v, nxt;
	ll w;
}e[N], re[N];
int h[N], rh[N], idx;

void AddEdge(int u, int v, ll w)
{
	e[++idx] = { v, h[u], w };
	h[u] = idx;
	re[idx] = { u, rh[v], w }; //反向边
	rh[v] = idx;
}

struct node
{
	int v;
	ll g, h; //s到v v到t  f = g + h
	bool operator < (const node &oth) const
	{
		return g + h > oth.g + oth.h;
	}
};
void Dijkstra()
{
	memset(dis, 0x3f, sizeof(dis));
	memset(vis, 0, sizeof(vis));
	dis[ed] = 0;
	priority_queue<node> pq;
	pq.push({ ed, 0, 0 });
	while (!pq.empty())
	{
		int u = pq.top().v, v;
		ll d = pq.top().h, w; pq.pop(); //存h内
		vis[u] = 1;
		for (int i = rh[u]; i; i = re[i].nxt) //跑反向边
		{
			v = re[i].v, w = re[i].w;
			if (!vis[v] && d + w < dis[v])
				dis[v] = d + w, pq.push({ v, 0, dis[v] });
		}
	}
}
ll AStar()
{
	Dijkstra(); //以ed为起点计算h值
	if (dis[st] == LINF) //无最短路
		return -1;
	priority_queue<node> pq;
	pq.push({ st, 0, 0 });
	while (!pq.empty())
	{
		int u = pq.top().v, v;
		ll d = pq.top().g, w; pq.pop(); //g
		if (u == ed && --k == 0) //是终点且是第k次到
			return d;
		for (int i = h[u]; i; i = e[i].nxt)
		{
			v = e[i].v, w = e[i].w;
			pq.push({ v, d + w, dis[v] });
		}
	}
	return -1;
}
int main()
{
#ifdef LOCAL
	//freopen("C:/input.txt", "r", stdin);
#endif
	while (cin >> n >> m)
	{
		idx = 0;
		memset(h, 0, sizeof(h));
		memset(rh, 0, sizeof(rh));
		cin >> st >> ed >> k >> t;
		for (int i = 1; i <= m; ++i)
		{
			int u, v;
			ll w;
			scanf("%d%d%lld", &u, &v, &w);
			AddEdge(u, v, w);
		}
		ll res = AStar();
		if (res == -1 || res > t)
			printf("Whitesnake!\n");
		else
			printf("yareyaredawa\n");
	}

	return 0;
}