B. Fire-Fighting Hero (dijstra优先队列+bfs)
题意:刚开始看错题了,以为是k次dijkstra,但是wa了,后来队友指正后发现挺水的。求S到其它点的最短路的最大值ans1,然后求其它点到指定k个点之一的最短路的最大值ans2。比较ans1和ans2即可。
思路:用dijstra优化队列求ans1,k次优先队列bfs求ans2即可。
AC code:
#include<cstdio> #include<algorithm> #include<cctype> #include<queue> using namespace std; inline int read() { int x=0,f=0; char ch=0; while(!isdigit(ch)) {f|=ch=='-';ch=getchar();} while(isdigit(ch)) x=(x<<3)+(x<<1)+(ch^48),ch=getchar(); return f?-x:x; } const int maxn=1005; const int maxm=5e5+5; const int inf=0x3f3f3f3f; int T,n,m,s,k,c,cnt,head[maxn],isk[maxn],dist[maxn],vis[maxn]; int ans1,ans2; struct node1{ int v,w,nex; }edge[maxm]; void adde(int u,int v,int w){ edge[++cnt].v=v; edge[cnt].w=w; edge[cnt].nex=head[u]; head[u]=cnt; } struct node2{ int w,id; node2(){} node2(int w,int id):w(w),id(id){} }; bool operator < (const node2& a,const node2& b){ return a.w>b.w; } void dijkstra(int s) { priority_queue<node2> que; for(int i=1; i<=n; i++) dist[i]=inf,vis[i]=0; dist[s]=0; que.push(node2(0,s)); while(!que.empty()) { int u=que.top().id; que.pop(); if(vis[u]) continue; vis[u]=1; for(int i=head[u];i;i=edge[i].nex) { int v=edge[i].v; int w=edge[i].w; if(dist[v]>dist[u]+w) { dist[v]=dist[u]+w; node2 rec; rec.id=v; rec.w=dist[v]; que.push(rec); } } } } int bfs(int s){ priority_queue<node2> que; for(int i=1; i<=n; ++i) vis[i]=0; que.push(node2(0,s)); while(!que.empty()){ node2 now=que.top();que.pop(); int nid=now.id,nw=now.w; if(vis[nid]) continue; vis[nid]=1; if(isk[nid]) return nw; for(int i=head[nid];i;i=edge[i].nex){ int v=edge[i].v; int w=edge[i].w; que.push(node2(nw+w,v)); } } } int main(){ T=read(); while(T--){ n=read(),m=read(),s=read(),k=read(),c=read(); cnt=0; ans1=ans2=0; for(int i=1;i<=n;++i) head[i]=0,isk[i]=0; for(int i=1;i<=k;++i) isk[read()]=1; for(int i=1;i<=m;++i){ int u=read(),v=read(),w=read(); adde(u,v,w); adde(v,u,w); } dijkstra(s); for(int i=1;i<=n;++i) ans1=max(ans1,dist[i]); for(int i=1;i<=n;++i){ if(isk[i]) continue; ans2=max(ans2,bfs(i)); } if(ans1<=c*ans2) printf("%d\n",ans1); else printf("%d\n",ans2); } return 0; }