思路:K短路模板,A*算法。估价函数=当前值+当前位置到终点的距离,即f(p)=g(p)+h(p),每次扩展估价函数值最小的一个。选择f(p)最小的点,若其为t,则计算其出队次数,次数等于k时,即为第k短路长度。
#include<cstdio>#include<queue>#include<cstring>#include<iostream>#include<algorithm>using namespace std;const int maxn=1e3+50;const int maxm=1e5+50;const int INF=0x3f3f3f3f;struct VNode{ int to,next,w;};struct Node{ int to,f,g; bool Operator <(const Node &rhs) const { if(rhs.f==f) return rhs.g<g; else return rhs.f<f; }};int S,T,K;int dist[maxn];int tot1,tot2,n,m;int head1[maxm],head2[maxm];VNode edge1[maxm],edge2[maxm];void addEdge1(int u,int v,int w){ edge1[tot1].to=v,edge1[tot1].w=w; edge1[tot1].next=head1[u],head1[u]=tot1++;}void addEdge2(int u,int v,int w){ edge2[tot2].to=v,edge2[tot2].w=w; edge2[tot2].next=head2[u],head2[u]=tot2++;}void SPFA(int s,int head[],VNode e[]){ int v[maxn]; queue<int> q; memset(v,0,sizeof(v)); for(int i=1; i<=n; i++) dist[i]=INF; v[s]=1,dist[s]=0,q.push(s); while(!q.empty()) { int now=q.front(); q.pop(),v[now]=0; for(int i=head[now]; ~i; i=e[i].next) { int nt=e[i].to; if(dist[nt]>dist[now]+e[i].w) { dist[nt]=dist[now]+e[i].w; if(!v[nt]) { v[nt]=1; q.push(nt); } } } }}int A_Star(int s,int t,int k,int head[],VNode e[]){ int cnt=0; if(s==t) k++; if(dist[s]==INF) return -1; Node now; now.to=s,now.g=0,now.f=now.g+dist[s]; priority_queue<Node> pq; pq.push(now); while(!pq.empty()) { Node now=pq.top(); pq.pop(); if(now.to==t) cnt++; if(cnt==k) return now.g; for(int i=head[now.to]; ~i; i=e[i].next) { Node nt; nt.to=e[i].to; nt.g=now.g+e[i].w; nt.f=nt.g+dist[nt.to]; pq.push(nt); } } return -1;}int main(){ while(scanf("%d%d",&n,&m)!=EOF) { tot1=0,tot2=0; memset(head1,-1,sizeof(head1)); memset(head2,-1,sizeof(head2)); for(int i=0; i<m; i++) { int x,y,w; scanf("%d%d%d",&x,&y,&w); addEdge1(x,y,w); addEdge2(y,x,w); } scanf("%d%d%d",&S,&T,&K); SPFA(T,head2,edge2); printf("%d/n",A_Star(S,T,K,head1,edge1)); } return 0;}
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