POJ 2449 Remmarguts' Date A* -

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求第s~t 第k短的路

先求出所有点到t的最短路径,这可以通过保存逆邻接表来求

这是求出h(u)的精确值dist[u],g()就是离s(起点)点经过多少距离

从优先队列里取出f最小的点,该点扩展数加1,扩展他连接的点,放入优先队列

原理类似Dijkstra,扩展到的点就是s出发最短路径,第i次扩展到u点就是到u点的第i短路径

#include<iostream>#include<cstdio>#include<vector>#include<queue>#include<algorithm>#include<cstring>using namespace std;typedef long long LL;const int maxn=1000+5;struct Point{int to,weight;Point(int t,int w):to(t),weight(w){}};vector<vector<Point> > G(maxn),GT(maxn);int n,m;int dist[maxn],inq[maxn];struct Node{int to;int f,g,h;Node(int to,int g):to(to),g(g){h=dist[to];f=g+h;}bool operator < (const Node& n) const {if(f==n.f) return g>n.g;return f>n.f;}};void Spfa(int s){memset(dist,0x6f,sizeof(dist));queue<int> Q;dist[s]=0;inq[s]=true;    Q.push(s);while(!Q.empty()){int u=Q.front(); Q.pop();inq[u]=false;for(int i=0;i<GT[u].size();i++){int v=GT[u][i].to, w=GT[u][i].weight;if(dist[v]>dist[u]+w){dist[v]=dist[u]+w;if(!inq[v]) Q.push(v),inq[v]=true;}}}}int closed[maxn];int A_Star(int s,int t,int k){if(dist[s]==dist[0]) return -1;if(s==t) k++;memset(closed,0,sizeof(closed));priority_queue<Node> open;open.push(Node(s,0));while(!open.empty()){int u=open.top().to,g=open.top().g; open.pop();closed[u]++;if(closed[t]==k) return g;for(int i=0;i<G[u].size();i++){int v=G[u][i].to,w=G[u][i].weight;open.push(Node(v,g+w));}}return -1;}int main(){int n,m,u,v,w,s,t,k;cin>>n>>m;while(m--){cin>>u>>v>>w;G[u].push_back(Point(v,w));GT[v].push_back(Point(u,w));}cin>>s>>t>>k;Spfa(t);cout<<A_Star(s,t,k)<<endl;return 0;}


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