Remmarguts’ Date

Description “Good man never makes girls wait or breaks an appointment!” said the mandarin duck father. Softly touching his little ducks’ head, he told them a story.

“Prince Remmarguts lives in his kingdom UDF – United Delta of Freedom. One day their neighboring country sent them Princess Uyuw on a diplomatic mission.”

“Erenow, the princess sent Remmarguts a letter, informing him that she would come to the hall and hold commercial talks with UDF if and only if the prince go and meet her via the K-th shortest path. (in fact, Uyuw does not want to come at all)”

Being interested in the trade development and such a lovely girl, Prince Remmarguts really became enamored. He needs you - the prime minister’s help!

DETAILS: UDF’s capital consists of N stations. The hall is numbered S, while the station numbered T denotes prince’ current place. M muddy directed sideways connect some of the stations. Remmarguts’ path to welcome the princess might include the same station twice or more than twice, even it is the station with number S or T. Different paths with same length will be considered disparate.

Input The first line contains two integer numbers N and M (1 <= N <= 1000, 0 <= M <= 100000). Stations are numbered from 1 to N. Each of the following M lines contains three integer numbers A, B and T (1 <= A, B <= N, 1 <= T <= 100). It shows that there is a directed sideway from A-th station to B-th station with time T. The last line consists of three integer numbers S, T and K (1 <= S, T <= N, 1 <= K <= 1000).

Output A single line consisting of a single integer number: the length (time required) to welcome Princess Uyuw using the K-th shortest path. If K-th shortest path does not exist, you should output “-1” (without quotes) instead.

Sample Input 2 2 1 2 5 2 1 4 1 2 2

Sample Output 14

題意：給出n個點m條邊的有向圖，求s到t的第k短路。

分析：估價函式f(x)，在這題中為x到t的最短路，即把反向圖求t到x的最短路；實際距離g(x)為f(x)+dist(x)，dist為s到x的第i短路。將g(x)放到一個堆中，每次取出g(x)最小的點，擴充套件與x相連的每條邊，直到第k次取出t，此時的dist(t)即為第k短路長度。

程式碼

#include <cstdio>
#include <queue>
#define N 200005
using namespace std;

struct arr
{
	int to, nxt, w;
}a[N],b[N];
struct node
{
	int x, dis, g;
	bool operator <(const node a)const
	{
		if (a.g == g) return a.dis < dis;
		return a.g < g;
	}
};
int n,m,s,t,k;
int d[N],lsa[N],la,lsb[N],lb;
bool vis[N];

void add(int x, int y, int z)
{
	a[++la].w = z;
	a[la].to = y;
	a[la].nxt = lsa[x];
	lsa[x] = la;
	b[++lb].w = z;
	b[lb].to = x;
	b[lb].nxt = lsb[y];
	lsb[y] = lb;
}

void spfa()
{
	queue<int> q;
	while (q.size()) q.pop();
	q.push(t);
	vis[t] = true;
	for (int i = 1; i <= n; i++) d[i] = 1e9;
	d[t] = 0;
	while (q.size())
	{
		int u = q.front();
		q.pop();
		vis[u] = false;
		for (int i = lsb[u]; i; i = b[i].nxt)
			if (d[b[i].to] > d[u] + b[i].w)
			{
				d[b[i].to] = d[u] + b[i].w;
				if (!vis[b[i].to]) 
				{
					vis[b[i].to] = true;
					q.push(b[i].to);
				}
			}
	}
}

int Astar()
{
	if (d[s] == 1e9) return -1;
	if (s == t) k++;
	int cnt = 0;
	priority_queue<node> Q;
	while (Q.size()) Q.pop();
	node u;
	u.x = s;
	u.dis = 0;
	u.g = d[s];
	Q.push(u);
	while (Q.size())
	{
		node v = Q.top();
		Q.pop();
		if (v.x == t)
		{
			cnt++;
			if (cnt == k) return v.dis;
		}
		for (int i = lsa[v.x]; i; i = a[i].nxt)
		{
			u.x = a[i].to;
			u.dis = v.dis + a[i].w;
			u.g = u.dis + d[u.x];
			Q.push(u);
		}
	}
	return -1;
}

int main()
{
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= m; i++)
	{
		int x, y, z;
		scanf("%d%d%d", &x, &y, &z);
		add(x, y, z);
	}
	scanf("%d%d%d", &s, &t, &k);
	spfa();
	printf("%d", Astar());
}