【K短路】【A星】Remmarguts' Date POJ 2449 A-Star

Remmarguts' Date

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions: 17296		Accepted: 4746

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

K短路裸体，第一次写K短路，WA了两次

每个点的 g 值不用说了，然后 h 值由当前点到终点 T 的最短路，所以要先把边反向，以T为源点来一次spfa

判断是否为K短路的时候应该在出队的时候判断，而不是入队的时候【应该很好理解，但是我一开始就错了。。。。】

然后这一题的S可能等于T，但一开始在S加一的时候并不算T加一，但实际上加了。。。。所以这种情况我们把他减去，或者直接把K+1，改成求K+1短路

测评情况（POJ）

第一次是判断K短路出错，第二次是没考虑S=T



C++ AC Code

/*http://blog.csdn.net/jiangzh7
By Jiangzh*/
#include<cstdio>
#include<cstring>
#include<queue>
using std::queue;
using std::priority_queue;
#define min(a,b) ((a)<(b)?(a):(b))
const int N=1000+10;
const int inf=0x3f3f3f3f;

int n,m;
struct Link{int y,z;Link *next;}*head
,*tail
;
int S,T,K;
int dist
,inQ
;
queue<int> que;
int enque_time
;

struct ND{
int x,g,h;
bool operator < (const ND &a) const
{
return g+h>a.g+a.h;
}
};
priority_queue<ND> Q;

void inhead(int x,int y,int z)
{
Link *p=new Link;
p->y=y;p->z=z;
p->next=head[x];
head[x]=p;
}
void intail(int x,int y,int z)
{
Link *p=new Link;
p->y=y;p->z=z;
p->next=tail[x];
tail[x]=p;
}

void read()
{
scanf("%d%d",&n,&m);
for(int i=1;i<=m;i++)
{
int x,y,z;
scanf("%d%d%d",&x,&y,&z);
inhead(x,y,z);
intail(y,x,z);
}
scanf("%d%d%d",&S,&T,&K);
}

bool spfa()
{
memset(dist,0x3f,sizeof(dist));
memset(inQ,0,sizeof(inQ));
dist[T]=0; que.push(T);
while(!que.empty())
{
int x=que.front();que.pop();
inQ[x]=0;
for(Link *p=tail[x];p;p=p->next)
if(dist[p->y]>dist[x]+p->z)
{
dist[p->y]=dist[x]+p->z;
if(!inQ[p->y])
{
inQ[p->y]=1;
que.push(p->y);
}
}
}
return dist[S]!=inf;
}

int solve()
{
if(!spfa()) return -1;
if(S==T) K++;
memset(enque_time,0,sizeof(enque_time));
ND start; start=(ND){S,0,dist[S]};
Q.push(start);
while(!Q.empty())
{
ND tmp=Q.top(); Q.pop();
int x=tmp.x;
enque_time[x]++;
if(enque_time[T]==K) return tmp.g;
for(Link *p=head[x];p;p=p->next)
{
ND tt=(ND){p->y,tmp.g+p->z,dist[p->y]};
Q.push(tt);
}
}
return -1;
}

int main()
{
freopen("2449.in","r",stdin);
freopen("2449.out","w",stdout);
read();
printf("%d\n",solve());
return 0;
}