poj 2449 Remmarguts' Date （k短路模板）

Remmarguts' Date

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions: 30772		Accepted: 8397

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

Sample Output

14

边可以重复走，
不严格的k短路
A*
估价函数为dis（起点，i）+dis（i，终点）

1、反向图上求出终点到每个点的最短路
2、起点入优先队列，
　　队首出队，
　 如果队首是终点，而且是第k次出队，
   那么当前距离就是k短路
  如果队首不是终点，便利与当前点连接的所有的点，入队

细节1：优先队列出入队不用vis数组判重，因为边可以重复走
细节2：如果第1步中，起点与终点不连通，输出-1结束，
　　　 否则进入A*，没有vis数组，出现环会死循环
细节3：如果起点=终点，令k++，因为起点会立即出队

#include<queue>
#include<cstdio>
#include<cstring>
#define N 1001
#define M 100001
using namespace std;
int n,s,t,k;
int dis1[N];
bool vis[N];
int front[N],to[M],nxt[M],val[M],tot;
int front2[N],to2[M],nxt2[M],val2[M],tot2;
struct node
{
    int num,dis;
    bool operator < (node p) const
    {
        return dis+dis1[num]>p.dis+dis1[p.num];
    } 
}now,nt;
void add(int u,int v,int w)
{
    to[++tot]=v; nxt[tot]=front[u]; front[u]=tot; val[tot]=w;
    to2[++tot2]=u; nxt2[tot2]=front2[v]; front2[v]=tot2; val2[tot2]=w; 
}
void init()
{
    int m,u,v,w;
    scanf("%d%d",&n,&m);
    while(m--)
    {
        scanf("%d%d%d",&u,&v,&w);
        add(u,v,w);
    }
    scanf("%d%d%d",&s,&t,&k);
}
void spfa()
{
    memset(dis1,63,sizeof(dis1));
    queue<int>q;
    dis1[t]=0;
    vis[t]=true;
    q.push(t);
    int now;
    while(!q.empty())
    {
        now=q.front();
        q.pop();
        vis[now]=false;
        for(int i=front2[now];i;i=nxt2[i])
            if(dis1[to2[i]]>dis1[now]+val2[i])
            {
                dis1[to2[i]]=dis1[now]+val2[i];
                if(!vis[to2[i]])
                {
                    q.push(to2[i]);
                    vis[to2[i]]=true;
                }
            }
    }
}
void Astar()
{
    if(dis1[s]>1e9) 
    {
        printf("-1");
        return;
    }    
    if(s==t) k++;
    int cnt=0,last=-1;
    priority_queue<node>q;
    now.num=s;
    now.dis=0;
    q.push(now);
    while(!q.empty())
    {
        now=q.top();
        q.pop();
        if(now.num==t) 
        {
            cnt++;
            if(cnt==k) 
            {
                printf("%d",now.dis);
                return;
            }
        }
        for(int i=front[now.num];i;i=nxt[i])
        {
            nt.num=to[i];
            nt.dis=now.dis+val[i];
            q.push(nt);
        }
    }
    printf("-1");
}
int main()
{
    init();
    spfa();
    Astar();
}