JavaScript实现Div二级联动效果（响应键盘按钮）

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;

int const N= 10010, inf= 0x7fffffff;
int n, k, cnt, tot, root, size, nchild, first, last, pre, ans;
int Fnum;

struct Edge{
int adj, wt;
Edge* next;
}tb[N<<1];

Edge* mat
;
int dist
, flag
, que
, vis
, num
;

void inline add( int u, int v, int d ){
++cnt;
tb[cnt].adj= v; tb[cnt].wt= d;
tb[cnt].next= mat[u]; mat[u]= tb+ cnt;

++cnt;
tb[cnt].adj= u; tb[cnt].wt= d;
tb[cnt].next= mat[v]; mat[v]= tb+ cnt;
}

int selectRoot( int rt ){
flag[rt]= Fnum;

int tmp= 0, sum= 0;
for( Edge* p= mat[rt]; p; p= p->next ){
int v= p->adj;

if( vis[v]== 0 && flag[v]!= Fnum ){
int s= selectRoot( v );

if( s> tmp ) tmp= s;
sum+= s;
}
}

if( nchild- 1- sum> tmp ) tmp= nchild- 1- sum;

if( tmp< size ){  size= tmp; root= rt; }

return sum+ 1;
}

int inline getCount( int first, int last ){
sort( dist+ first, dist+ last+ 1 );

int head= first, tail= last, ret= 0;
while( first< last ){
if( dist[first]+ dist[last]<= k ) ret+= last- first++;
else last--;
}
return ret;
}

int dfs( int rt, int len ){
flag[rt]= Fnum; dist[++last]= len;

int sum= 0;
for( Edge* p= mat[rt]; p; p= p->next )
if( !vis[p->adj] && flag[p->adj]!= Fnum ){
int v= p->adj, wt= p->wt;

sum+= dfs( v, len+ wt );
}

num[rt]= sum+ 1;
return sum+ 1;
}

void solve( int rt ){
pre= 1;
for( Edge* p= mat[rt]; p; p= p->next )
if( !vis[p->adj] && flag[p->adj]!= Fnum ){
int v= p->adj, wt= p->wt;

dfs( v, wt );
ans-= getCount( pre, last );

pre= last+ 1;
}
}

int main(){
while( scanf("%d%d",&n,&k)!= EOF ){
if( n== 0 && k== 0 ) return 0;

for( int i= 0; i<= n; ++i ) mat[i]= 0, flag[i]= 0, vis[i]= 0;
cnt= -1; tot= -1;

int u, v, d;
for( int i= 1; i< n; ++i ){
scanf("%d%d%d", &u,&v,&d );

add( u, v, d );
}

int head= 0, tail= 0; que[0]= 1;
nchild= n; ans= 0; Fnum= 0; num[1]= n;

while( head<= tail ){
int now= que[head++]; nchild= num[now]; Fnum++;

size= 0x7fffffff; selectRoot( now ); vis[root]= 1;

for( Edge*p= mat[root]; p; p= p->next )
if( !vis[p->adj] ) que[++tail]= p->adj;

first= pre= 1; last= 0; Fnum++;

solve( root ); dist[++last]= 0;
ans+= getCount( first, last );
}

printf("%d\n", ans );
}

return 0;
}