题解:建出分层图,跑最短路
经验教训:一定要检查空间,并不都是开两倍的m!!!!!!!!!!!!!!!!!!!!!!!!!!!
仔细检查是否多开了一个0或少开了一个0
#include<iostream> #include<cstdio> #include<cstring> #include<queue> using namespace std; const int maxm=50009; const int maxn=10009; const int maxk=13; const int oo=1000000000; int n,m,k; int p[11][maxn]; int totn=0; int ans=oo; int cntedge=0; int head[maxn*maxk]={0}; int to[maxm*maxk*4]={0},nex[maxm*maxk*4]={0},dist[maxm*maxk*4]={0}; void Addedge(int x,int y,int z){ nex[++cntedge]=head[x]; to[cntedge]=y; dist[cntedge]=z; head[x]=cntedge; } int d[maxn*maxk]; int vis[maxn*maxk]; struct HeapNode{ int v,mindist; HeapNode(int x){ v=x;mindist=d[x]; } bool operator < (const HeapNode &rhs) const{ return mindist>rhs.mindist; } }; priority_queue<HeapNode>q; int s,t; void Dijkstra(){ for(int i=1;i<=totn;++i)d[i]=oo; d[p[0][s]]=0;q.push(HeapNode(p[0][s])); while(!q.empty()){ HeapNode x=q.top();q.pop(); int u=x.v; if(vis[u])continue; vis[u]=1; for(int i=head[u];i;i=nex[i]){ if(d[u]+dist[i]<d[to[i]]){ d[to[i]]=d[u]+dist[i]; q.push(HeapNode(to[i])); } } } } int main(){ scanf("%d%d%d",&n,&m,&k); for(int j=0;j<=k;++j){ for(int i=1;i<=n;++i){ p[j][i]=++totn; } } scanf("%d%d",&s,&t); ++s;++t; while(m--){ int x,y,z; scanf("%d%d%d",&x,&y,&z); ++x;++y; for(int i=0;i<=k;++i){ Addedge(p[i][x],p[i][y],z); Addedge(p[i][y],p[i][x],z); } for(int i=0;i<k;++i){ Addedge(p[i][x],p[i+1][y],0); Addedge(p[i][y],p[i+1][x],0); } } Dijkstra(); for(int i=0;i<=k;++i)ans=min(ans,d[p[i][t]]); printf("%d\n",ans); return 0; }