Remmarguts‘ Date
| Time Limit: 4000MS | Memory Limit: 65536K | |
| Total Submissions: 21084 | Accepted: 5740 |
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
该题意所给的是有向边。求的是从起点S到终点T的第K最短路。
解题：因为A*算法是以最快的方法求出一条最短路，所以这里第 i 次算出一条路到终点那么一定是第i最短路。
那么我们现在的任务关建就是设计A*中的估价F[i],而F[i]=G[i]+H[i],G[i]表示的是从起始点到当前点i的路径总和，这个可以在过程中求出；H[i]表示的是从当前点到终点的路径总和，每次从开放集合中取出最小F[i]的点，这个可以用优先队列表示开放集合。明白这里之后，我们的任务就是要算出每个点到终点的H值，那么这样我们可以建一个反向图，从起点到达每一个点的最短路也就算完成了求出每个点的H值。
注意两点：1.每个点我们最多只求出K条路，有大于K条路就不必再往下走了，因为当前点的第K+1条最短路到达终点时一定不在终点的K条最短路之列。
2.有的点是走不到终点的，那么我们就不用考虑这一类的点，可以列出黑名单里，而这个判断就是通过H[i]值。
这样我们便顺利的完成了这一类题。
#include<stdio.h>
#include<iostream>
#include<queue>
#include<vector>
using namespace std;
#define inf 99999999
#define N 1100
typedef struct nnn
{
int F,G,s;
friend bool operator<(nnn a,nnn b)
{
return a.F>b.F;
}
}PATH;
typedef struct nn
{
int v,w;
}node;
vector<node>map[N],tmap[N];
int H[N];
void findH(int s)
{
queue<int>q;
int inq[N]={0};
q.push(s); inq[s]=1; H[s]=0;
while(!q.empty())
{
s=q.front(); q.pop(); inq[s]=0;
int m=tmap[s].size();
for(int i=0;i<m;i++)
{
int j=tmap[s][i].v;
if(H[j]>tmap[s][i].w+H[s])
{
H[j]=tmap[s][i].w+H[s];
if(!inq[j])
inq[j]=1,q.push(j);
}
}
}
}
int Astar(int st,int end,int K)
{
priority_queue<PATH>q;
PATH p,tp;
int k[N]={0};
findH(end);
if(H[st]==inf)return -1;
p.s=st; p.G=0; p.F=H[st];
q.push(p);
while(!q.empty())
{
p=q.top(); q.pop();
k[p.s]++;
if(k[p.s]>K)continue;//每个点最多走K次，超过K条路不必走
if(p.s==end&&k[end]==K) return p.F;
int m=map[p.s].size();
for(int i=0;i<m;i++)
{
int j=map[p.s][i].v;
if(H[j]!=inf)//表明当前点不能通向终点，就不用加入队列
{
tp.G=p.G+map[p.s][i].w;
tp.F=H[j]+tp.G;
tp.s=j;
q.push(tp);
}
}
}
return -1;
}
int main()
{
int n,m,S,T,K,a,b,t;
node p;
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
{
map[i].clear(); tmap[i].clear(); H[i]=inf;
}
while(m--)
{
scanf("%d%d%d",&a,&b,&t);
p.v=b; p.w=t; map[a].push_back(p);
p.v=a; tmap[b].push_back(p);//反建一个地图求H
}
scanf("%d%d%d",&S,&T,&K);
if(S==T)K++;
printf("%d\n",Astar(S,T,K));
}
POJ2449Remmarguts' Date（A*算法求第K小路）