[树的直径 树形DP] UOJ #11【UTR #1】ydc的大树

题解链接

大意是我们求出直径的中心 那么到最远点必然经过中心
把它提到根
就可以在枚举删的白点时很快统计出能够使多少个黑点不高兴

#include<cstdio>#include<cstdlib>#include<algorithm>using namespace std;inline char nc(){  static char buf[100000],*p1=buf,*p2=buf;  return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;}inline void read(int &x){  char c=nc(),b=1;  for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;  for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;}const int N=100005;struct edge{  int u,v,w,next;}G[N<<1];int head[N],inum;inline void add(int u,int v,int w,int p){  G[p].u=u; G[p].v=v; G[p].w=w; G[p].next=head[u]; head[u]=p;}#define V G[p].vint bla[N];int dis[N],dis2[N];int pt,maxd;inline void find(int u,int fa){  if (bla[u] && dis[u]>maxd) maxd=dis[u],pt=u;  for (int p=head[u];p;p=G[p].next)    if (V!=fa)      dis[V]=dis[u]+G[p].w,find(V,u);}int size[N];int cur,rt[N];int f[N],g[N];inline void dfs(int u,int fa){  rt[u]=cur;  size[u]=bla[u];  for (int p=head[u];p;p=G[p].next)    if (V!=fa)      dis[V]=dis[u]+G[p].w,dfs(V,u),size[u]+=size[V];  if (bla[u]) f[u]=dis[u],g[u]=1;  for (int p=head[u];p;p=G[p].next)    if (V!=fa){      if (f[V]>f[u])    f[u]=f[V],g[u]=g[V];      else if (f[V]==f[u])    g[u]+=g[V];    }}int n,m,RT;int main(){  int x,y,iu,iv,iw;  freopen("t.in","r",stdin);  freopen("t.out","w",stdout);  read(n); read(m); y=m;  while (y--) read(x),bla[x]=1;  for (int i=1;i<n;i++)    read(iu),read(iv),read(iw),add(iu,iv,iw,++inum),add(iv,iu,iw,++inum);  maxd=0; find(1,0);  for (int i=1;i<=n;i++) dis[i]=0;  x=pt; maxd=0; find(pt,0);  for (int i=1;i<=n;i++) dis2[i]=dis[i],dis[i]=0;;  maxd=0; find(pt,0);  RT=0;  for (int i=1;i<=n;i++)    if(dis[i]+dis2[i]==dis[x])      if(!RT || abs(dis[i]-dis2[i])<abs(dis[RT]-dis2[RT]))    RT=i;  dis[RT]=0,size[RT]=bla[RT];  for (int p=head[RT];p;p=G[p].next)    dis[V]=G[p].w,cur=V,dfs(V,RT),size[RT]+=size[V];  int mx1=0,mx2=0,id1=0,id2=0;  for(int p=head[RT];p && ~id1 && ~id2;p=G[p].next){    if (!size[V]) continue;    if (f[V]>mx1)      mx2=mx1,id2=id1,mx1=f[V],id1=V;    else if(f[V]==mx1)      mx1==mx2?id1=id2=-1:(mx2=f[V],id2=V);    else if(f[V]>mx2)      mx2=f[V],id2=V;  }  int ans=0,tot=0;  if (!bla[RT]) ans=m,tot=1;  for (int i=1;i<=n;i++){    if (i==RT || bla[i]) continue;    int ret=size[i];    if (~id1 && ~id2 && size[i] && f[i]==f[rt[i]] && g[i]==g[rt[i]]){      if (rt[i]==id1)        f[id1]>f[id2]?ret+=m-size[id1]:ret+=size[id2];      else if (rt[i]==id2)    ret+=size[id1];    }    if (ret>ans)      ans=ret,tot=1;    else if(ret==ans)      tot++;  }  printf("%d %d\n",ans,tot);  return 0;}