POJ 3728 The merchant

这个题算是LCA的一个高级应用吧。

其实就是在维护u,v到其最近公共祖先的几个数据（将u,v分成两段,一个从u到LCA(u,v),一个从LCA(u,v)到v,分别维护即可）

#include<cstdio> #include<string.h> #include<math.h> #define con 50100 using namespace std; int head1[con],head2[con],head3[con],p,n,num[con],mi[con],mx[con],up[con],down[con],vis[con],f[50100],ans[con]; struct Edge{ int f; int to; int d; int next; }edge[con*4]; int max(int a,int b){ return a>b?a:b; } int min(int a,int b){ return a>b?b:a; } int find(int u){ //这里用递归与分治更新数据 if(f[u]==u)return u; int tem=f[u]; f[u]=find(f[u]); up[u]=max(max(up[u],up[tem]),mx[tem]-mi[u]); down[u]=max(max(down[u],down[tem]),mx[u]-mi[tem]); mx[u]=max(mx[u],mx[tem]); mi[u]=min(mi[u],mi[tem]); return f[u]; } void tarjan(int u){ int i,j; f[u]=u; for(i=head1[u];i;i=edge[i].next){ int v=edge[i].to; if(!vis[v]){ tarjan(v); f[v]=u; } } vis[u]=1; for(i=head2[u];i;i=edge[i].next){ int v=edge[i].to; if(vis[v]){ int t=find(v); edge[++p].d=i; edge[p].next=head3[t],head3[t]=p; } } //根据u,v的LCA分类 for(i=head3[u];i;i=edge[i].next){ int kk=edge[i].d; int t=edge[kk].d,x=edge[kk].f,y=edge[kk].to; find(x); if(t<0){ int tem=x; x=y,y=tem; t=-t; } ans[t]=max(max(up[x],down[y]),mx[y]-mi[x]); } } int main(){ int i,j,m; scanf("%d",&n); for(i=1;i<=n;i++){ scanf("%d",&num[i]); mi[i]=mx[i]=num[i]; up[i]=0,down[i]=0; } memset(head1,0,sizeof(head1)); memset(head2,0,sizeof(head2)); memset(vis,0,sizeof(vis)); for(i=1;i<n;i++){ int u,v; scanf("%d %d",&u,&v); edge[++p].to=v; edge[p].next=head1[u],head1[u]=p; } scanf("%d",&m); for(i=1;i<=m;i++){ int u,v; scanf("%d %d",&u,&v); edge[++p].to=v; edge[p].f=u,edge[p].d=i,edge[p].next=head2[u],head2[u]=p; edge[++p].to=u; edge[p].f=v,edge[p].d=-i,edge[p].next=head2[v],head2[v]=p; } tarjan(1); for(i=1;i<=m;i++){ printf("%d\n",ans[i]); } }