题解【AcWing178】第K短路

题面

经典的 ( ext{A*}) 算法例题。

我们可以把估价函数 (f(i)) 设置成当前点 (i) 到终点 (t) 的最短距离，这可以建反图使用 ( ext{Dijkstra}) 算法预处理出。

然后一遍 ( ext{BFS}) 即可。

注意当起点与终点相同时要 (++k) 。

#include <bits/stdc++.h>
#define DEBUG fprintf(stderr, "Passing [%s] line %d
", __FUNCTION__, __LINE__)
#define itn int
#define gI gi

using namespace std;

typedef pair <int, int> PII;
typedef pair <int, PII> PIII;

inline int gi()
{
	int f = 1, x = 0; char c = getchar();
	while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
	while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
	return f * x;
}

const int maxn = 1003, maxm = 100003;

int s, t, k, n, m;
int tot, head[maxn], ver[maxm * 2], nxt[maxm * 2], edge[maxm * 2], headu[maxn];
int dist[maxn], f[maxn], g[maxn], vis[maxn];

inline void add(int *h, int u, int v, int w)
{
	ver[++tot] = v, edge[tot] = w, nxt[tot] = h[u], h[u] = tot;
}

inline void Dijkstra() //预处理估价函数值
{
	priority_queue <PII, vector <PII>, greater <PII> > q;
	memset(dist, 0x3f, sizeof(dist));
	dist[t] = 0;
	q.push(make_pair(0, t));
	while (!q.empty())
	{
		int u = q.top().second; q.pop();
		if (vis[u]) continue;
		vis[u] = 1;
		for (int i = headu[u]; i; i = nxt[i])
		{
			int v = ver[i], w = edge[i];
			if (dist[v] > dist[u] + w) 
			{
				dist[v] = dist[u] + w;
				q.push(make_pair(dist[v], v));
			} 
		}
	}
	memcpy(f, dist, sizeof(f));
}

inline int A_star()
{
	priority_queue <PIII, vector <PIII>, greater <PIII> > q;
	//优先队列中存储 3 个信息：估价函数值、实际距离与当前节点编号
	memset(vis, 0, sizeof(vis));
	q.push(make_pair(f[s], make_pair(0, s))); //加入起点
	while (!q.empty())
	{
		int now = q.top().second.second, dis = q.top().second.first;
		q.pop();
		if (vis[now] >= k) continue; //已经搜过了 K 短路
		++vis[now];
		if (now == t && vis[now] == k) return dis; //找到了结果
		for (int i = head[now]; i; i = nxt[i])
		{
			int v = ver[i], w = edge[i];
			if (vis[v] < k)
			{
				q.push(make_pair(f[v] + dis + w, make_pair(dis + w, v))); //进行扩展
			}
		}
	}
	return -1;
}

int main()
{
	//freopen(".in", "r", stdin);
	//freopen(".out", "w", stdout);
	n = gi(), m = gi();
	for (int i = 1; i <= m; i+=1)
	{
		int u = gi(), v = gi(), w = gi();
		add(head, u, v, w), add(headu, v, u, w); //建反图
	}
	s = gi(), t = gi(), k = gi();
	if (s == t) ++k;
	Dijkstra();
	printf("%d
", A_star());
	return 0;
}