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
這道題是給出了一個圖，然後給出起點s與終點t與k，要求s到t的第k短路是多少
首先這肯定是一個最短路的問題，我們可以用SPFA算出t到其他所有點的最短路徑，然後使用A*去迭代搜索，一開始，把s到t的最短路先壓入隊列中，那麼就變成了以t為起點去遍歷，在第k次回到t點的時候所走過的路徑最短的值便是第k短路徑
#include
#include
#include
#include
using namespace std;
const int L = 100005;
const int inf = 1<<30;
struct node
{
int now,g,f;
bool operator <(const node a)const
{
if(a.f == f) return a.g < g;
return a.f < f;
}
};
struct edges
{
int x,y,w,next;
} e[L<<2],re[L<<2];//e是輸入給出的定向圖，re為其逆向圖，用於求t到其他所有點的最短路徑
int head[1005],dis[1005],vis[1005],n,m,k,s,t,rehead[1005];
void Init()//初始化
{
memset(e,-1,sizeof(e));
memset(re,-1,sizeof(re));
for(int i = 0; i<=n; i++)
{
dis[i] = inf;
vis[i] = 0;
head[i] = -1;
rehead[i] = -1;
}
}
void AddEdges(int x,int y,int w,int k)
{
e[k].x = x,e[k].y = y,e[k].w = w,e[k].next = head[x],head[x] = k;
re[k].x = y,re[k].y = x,re[k].w = w,re[k].next = rehead[y],rehead[y] = k;
}
int relax(int u,int v,int c)
{
if(dis[v]>dis[u]+c)
{
dis[v] = dis[u]+c;
return 1;
}
return 0;
}
void SPFA(int src)
{
int i;
dis[src] = 0;
queue Q;
Q.push(src);
while(!Q.empty())
{
int u,v;
u = Q.front();
Q.pop();
vis[u] = 0;
for(i = rehead[u]; i!=-1; i = re[i].next)
{
v = re[i].y;
if(relax(u,v,re[i].w) && !vis[v])
{
Q.push(v);
vis[v] = 1;
}
}
}
}
int Astar(int src,int to)
{
priority_queue Q;
int i,cnt = 0;
if(src == to) k++;//在起點與終點是同一點的情況下，k要+1
if(dis[src] == inf) return -1;
node a,next;
a.now = src;
a.g = 0;
a.f = dis[src];
Q.push(a);
while(!Q.empty())
{
a = Q.top();
Q.pop();
if(a.now == to)
{
cnt++;
if(cnt == k)
return a.g;
}
for(i = head[a.now]; i!=-1; i = e[i].next)
{
next = a;
next.now = e[i].y;
next.g = a.g+e[i].w;
next.f = next.g+dis[next.now];
Q.push(next);
}
}
return -1;
}
int main()
{
int i,j,x,y,w;
while(~scanf("%d%d",&n,&m))
{
Init();
for(i = 0; i
