HDU2449 Remmarguts' Date(第k短路)

Remmarguts' Date

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions: 24007		Accepted: 6535

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

题目大意:

给一个图，求第k短路.

解题思路:

运用spfa反向求一次最短路，可以求得终点(反向后变成起点)到各点的最短路径，变成A*的估价函数，A*的方程为f(v) = g(v) + h(v),其中f为估价函数，g(v)为出发点到该点已经走的距离，h(v)为预处理的距离，也就是刚才spfa反向求得起点到各点的最短路值。

AC代码:

#include
#include
#include 
#include
using namespace std;

const int maxe = 100000;
const int inf = 0x3f3f3f3f;
struct Node1
{
	int to;
	int g,f; // f = g + h;
	bool operator < (const Node1 &r) const
	{
		if(r.f == f) return r.g < g;
		return r.f < f;	
	}	
};

int dist[maxe];
int head[maxe];
int ehead[maxe];
int outque[maxe];

struct Node
{
	int to;
	int w;
	int next;
}edge[maxe*3],edge1[maxe*3];

int cnt1,cnt2;

void init()
{
	cnt1 = 0;
	cnt2 = 0;
	memset(head,-1,sizeof(head));
	memset(ehead,-1,sizeof(ehead));	
}

void addEdge(int u,int v,int w)
{
	edge[cnt1].to = v;
	edge[cnt1].w = w;
	edge[cnt1].next = head[u];
	head[u] = cnt1++; 
}

void add_Edge(int u,int v,int w)
{
	edge1[cnt2].to = v;
	edge1[cnt2].w = w;
	edge1[cnt2].next = ehead[u];
	ehead[u] = cnt2++;
}

bool SPFA(int s,int n)
{
	int i,k;
	bool vis[maxe];
	int queue[maxe];
	int iq;
	int top;
	for(i=0;i<=n;i++)
	{
		dist[i] = inf;
	}
	memset(vis,false,sizeof(vis));
	memset(outque,0,sizeof(outque));
	iq = 0;
	queue[iq++] = s;
	vis[s] = true;
	dist[s] = 0;
	i = 0;
	while(i != iq)
	{
		top = queue[i];
		vis[top] = false;
		outque[top]++;
		if(outque[top] > n) return false;
		k = ehead[top];
		while(k >= 0)
		{
			if(dist[edge1[k].to] - edge1[k].w > dist[top])
			{
				dist[edge1[k].to] = dist[top] + edge1[k].w;
				if(!vis[edge1[k].to])
				{
					vis[edge1[k].to] = true;
					queue[iq] = edge1[k].to;
					iq++;
				}
			}
			k = edge1[k].next;
		}
		i++;
	}
	return true;
}

int a_star(int start,int end,int n,int k)
{
	Node1 e,ne;
	int cnt = 0;
	priority_queue que;
	if(start == end) k++; //起点终点一样+1
	if(dist[start] == inf)
	{
		return -1;	
	}	
	e.to = start;
	e.g = 0;
	e.f = e.g + dist[e.to];
	que.push(e);
	while(!que.empty())
	{
		e = que.top();
		que.pop();
		if(e.to == end)
		{
			cnt++;
		}
		if(cnt == k)
		{
			return e.g;
		}
		for(int i=head[e.to];i != -1;i = edge[i].next)
		{
			ne.to = edge[i].to;
			ne.g = e.g + edge[i].w;
			ne.f = ne.g + dist[ne.to];
			que.push(ne);
		}
	}
	return -1;
}

int main()
{
	int m,n;
	int i;
	int a,b,time;
	int st,ed,ki;
	//freopen("1.txt","r",stdin);
	while(scanf("%d%d",&m,&n) != EOF)
	{
		init();
		for(i=1;i<=n;i++)
		{
			scanf("%d%d%d",&a,&b,&time);
			addEdge(a,b,time);
			add_Edge(b,a,time);	
		}
		scanf("%d%d%d",&st,&ed,&ki);
		SPFA(ed,m); //反向终点变成起点 
		printf("%d\n",a_star(st,ed,m,ki));
	}
	return 0;
}