虚拟赛的时候,一看是求任意两点的最短路,不管怎么优化都超时,最近做一些树的题目,这是求任意两点到最近公共祖先的距离,先用并差集判断是否联通,联通就求最近公共祖先,先把图建成一棵棵树,
#include#include #define N 10010 int father[N],dfs[N],n,vis[N],head[N],num,f[N],ans,dis[N]; struct edge { int st,ed,next,w; }E[N*2]; void addedge(int x,int y,int w) { E[num].st=x; E[num].ed=y; E[num].w=w; E[num].next=head[x]; head[x]=num++; } int find(int a) { if(f[a]!=a) f[a]=find(f[a]); return f[a]; } void dfs1(int u) { int i,v; vis[u]=1; for(i=head[u];i!=-1;i=E[i].next) { v=E[i].ed; if(vis[v]==1)continue; father[v]=u; dfs[v]=dfs[u]+1; dis[v]=E[i].w; dfs1(v); } if(ans dfs[y]) { sum+=dis[x]; x=father[x]; } while(dfs[y]>dfs[x]) { sum+=dis[y]; y=father[y]; } while(x!=y) { sum+=(dis[x]+dis[y]); x=father[x]; y=father[y]; } return sum; } int main() { int i,j,k,x,y,m,ans,w; while(scanf("%d%d%d",&n,&m,&k)!=-1) { memset(head,-1,sizeof(head)); num=0; for(i=1;i<=n;i++) f[i]=i; for(i=0;i