Remmarguts' Date
| Time Limit: 4000MS | Memory Limit: 65536K | |
| Total Submissions: 37028 | Accepted: 10206 |
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
題意：求第K短路，每個點可以重複訪問。
思路：A*算法的應用。A*是一種求解最短路的算法。
設源點爲s，目標點爲t，當前點爲x；
g(x)：表示從s到達x的實際距離；
h(x)：表示從x到達t的估計距離；
f(x)=g(x)+h(x)；
表示按照到達x點的走法從s到達t的估計距離（實際距離+估計距離=估計距離）。
因此我們需要先求得每一個點x到目標點t的距離即h(x);建立反向邊，使用任意的最短路徑算法即可求的h(x)
A*算法的本質是BFS，使用優先隊列目的是每次取的f（x）爲最小。因此每次出隊的點如果是t。則對應的f(x)的就爲最短距離。出隊k次，則對應的f(x)爲k短路。
AC代碼：
A*（圖的搜索方法）+spfa（計算h(x)）+鏈式前向星（存邊）
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
using namespace std;
const int MAXN = 1e3+5;
const int MAXM = 5e5+5;
const int INF = 0x3f3f3f3f;
struct A{
int f,g,v;
//從小到大排序
bool operator <(const A a)const{
if(a.f==f) return a.g<g;
return a.f<f;
}
};
priority_queue<A> q2;
struct Edge{
int next,v,w;
}edge[MAXM],revedge[MAXM];
int cnt,vis[MAXN],d[MAXN],head[MAXM],revhead[MAXM];
queue<int> q1;
void add(int u,int v,int w){
edge[cnt]=(Edge){head[u],v,w};
head[u]=cnt++;
revedge[cnt]=(Edge){revhead[v],u,w};
revhead[v]=cnt++;
}
void spfa(int s){
fill(d,d+MAXN,INF);
memset(vis,0,sizeof(vis));
d[s]=0;
q1.push(s);
while(!q1.empty()){
int u = q1.front();
q1.pop();
vis[u]=false;
for(int i=revhead[u];~i;i=revedge[i].next){
int v=revedge[i].v;
int w=revedge[i].w;
if(d[v]>d[u]+w){
d[v]=d[u]+w;
if(!vis[v]){
vis[v]=true;
q1.push(v);
}
}
}
}
}
int Astar(int s,int t,int k){
int tmp=0;//記錄終點第幾次出隊
while(!q2.empty()){
q2.pop();
}
/*如果起點與終點爲同一點
，不移動不算爲一條路徑，
然而我們卻多統計了這一條，故k++；*/
if(s==t) k++;
//圖不是連通圖
if(d[s]==INF) return -1;
A x;
x.v=s;x.g=0;x.f=x.g+d[s];
q2.push(x);
while(!q2.empty()){
A now = q2.top();
q2.pop();
if(now.v==t){
tmp++;
if(tmp==k) return now.g;
}
for(int i=head[now.v];~i;i=edge[i].next){
x.v=edge[i].v;
x.g=now.g+edge[i].w;
x.f=x.g+d[x.v];
q2.push(x);
}
}
return -1;
}
int main(){
int n,m;
while(~scanf("%d%d",&n,&m)){
memset(head,-1,sizeof(head));
memset(revhead,-1,sizeof(revhead));
cnt=0;
for(int i=0;i<m;i++){
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
add(u,v,w);
}
int s,t,k;
scanf("%d%d%d",&s,&t,&k);
spfa(t);
printf("%d\n",Astar(s,t,k));
}
return 0;
}