BZOJ 2152 Cong Coco (tree Division)

BZOJ 2152 Cong Coco (tree Division)

Address: BZOJ 2152
Find the number of vertices with the sum of three. The simplest point grouping.
The Code is as follows:

#include 
  
   #include 
   
    #include 
    
     #include 
     
      #include #include 
      
       #include
       #include 
        
         #include 
         
          #include 
          
           using namespace std;#define LL __int64#define pi acos(-1.0)//#pragma comment(linker, /STACK:1024000000)const int mod=1e9+7;const int INF=0x3f3f3f3f;const double eqs=1e-9;const int MAXN=20000+10;int head[MAXN], cnt, root, ans, min1;int siz[MAXN], ha[4], dep[MAXN], vis[MAXN];struct node{ int v, w, next;}edge[MAXN<<1];void add(int u, int v, int w){ edge[cnt].v=v; edge[cnt].w=w; edge[cnt].next=head[u]; head[u]=cnt++;}void init(){ memset(head,-1,sizeof(head)); cnt=ans=0; memset(vis,0,sizeof(vis));}void getroot(int u, int fa, int s){ siz[u]=1; int i, max1=0; for(i=head[u];i!=-1;i=edge[i].next){ int v=edge[i].v; if(v==fa||vis[v]) continue ; getroot(v,u,s); siz[u]+=siz[v]; max1=max(max1,siz[v]); } max1=max(max1,s-siz[u]); if(min1>max1){ root=u; min1=max1; }}void getdep(int u, int fa){ ha[dep[u]%3]++; siz[u]=1; for(int i=head[u];i!=-1;i=edge[i].next){ int v=edge[i].v; if(v==fa||vis[v])continue ; dep[v]=dep[u]+edge[i].w; getdep(v,u); siz[u]+=siz[v]; }}int Cal(int u, int len){ dep[u]=len; ha[0]=ha[1]=ha[2]=0; getdep(u,-1); return ha[0]*ha[0]+ha[1]*ha[2]*2;}void work(int u){ vis[u]=1; ans+=Cal(u,0); for(int i=head[u];i!=-1;i=edge[i].next){ int v=edge[i].v; if(vis[v]) continue ; ans-=Cal(v,edge[i].w); min1=INF; getroot(v,-1,siz[v]); work(root); }}int gcd(int x, int y){ return x==0?y:gcd(y%x,x);}int main(){ int n, u, v, w, i, j, _gcd; while(scanf(%d,&n)!=EOF){ init(); for(i=1;i