POJ 2447 Remmarguts' Date【k短路 SPFA+A* 模板题】

Remmarguts' Date

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions: 26059		Accepted: 7079

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

Source

POJ Monthly,Zeyuan Zhu

第k短路，存图时，正向反向都存，然后根据反向图spfa跑一遍，因为之后用A* 做铺垫，恩，A* ，通过估价函数确定当前点到终点的距离，决定搜索方向，这里估价函数=当前值+当前位置到终点的距离，f(p)=h(p)+g(p) ，每次扩展估价函数值最小的一个。对于k 短路， g(p) 为当前从起点到当前点的路径长度，h(p)为从当前点到终点的最短路的长度（之前spfa求得的距离）

#include <iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#define maxn 1010
#define inf 0x3f3f3f3f
using namespace std;
int head[maxn],rhead[maxn];
int n,k,s,e,cnt,dis[maxn];
bool vis[maxn];
struct node
{
int f,t,w,next;
};
struct rnode
{
int t,g,f;
bool operator < (const rnode r)const
{
if(r.f==f)
return r.g < g;
else
return r.f < f;
}
};
node edge[maxn*100],redge[maxn*100];
void add(int f,int t,int w)
{
edge[cnt].f=f;
edge[cnt].t=t;
edge[cnt].w=w;
edge[cnt].next=head[f];
head[f]=cnt;
redge[cnt].f=t;
redge[cnt].t=f;
redge[cnt].w=w;
redge[cnt].next=rhead[t];
rhead[t]=cnt++;
}
bool spfa()
{
int i,k;
memset(vis,false,sizeof(vis));
queue<int>q;
for(int i=0;i<=n;++i)
dis[i]=inf;
vis[e]=true;
dis[e]=0;
q.push(e);
while(!q.empty())
{
int v=q.front();
q.pop();
vis[v]=false;
for(int i=rhead[v];i!=-1;i=redge[i].next)
{
int u=redge[i].t;
int w=redge[i].w;
if(dis[u]>dis[v]+w)
{
dis[u]=dis[v]+w;
if(!vis[u])
{
vis[u]=true;
q.push(u);
}
}
}
}
return true;
}
int a_star()
{
rnode now,next;
int num=0;
priority_queue<rnode>q;
if(s==e)
k++;
if(dis[s]==inf)
return -1;
now.t=s;
now.g=0;
now.f=now.g+dis[s];
q.push(now);
while(!q.empty())
{
now=q.top();
q.pop();
if(now.t==e)
num++;
if(num==k)
return now.g;
for(int i=head[now.t];i!=-1;i=edge[i].next)
{
next.t=edge[i].t;
next.g=now.g+edge[i].w;
next.f=next.g+dis[next.t];
q.push(next);
}
}
return -1;
}
int main()
{
int m,a,b,t;
while(~scanf("%d%d",&n,&m))
{
cnt=0;
memset(head,-1,sizeof(head));
memset(rhead,-1,sizeof(rhead));
for(int i=0;i<m;++i)
{
scanf("%d%d%d",&a,&b,&t);
add(a,b,t);
}
scanf("%d%d%d",&s,&e,&k);
spfa();
int klength=a_star();
printf("%d\n",klength);
}
return 0;
}