hdu5016

题意：给定一个n个点的图，这个图是一棵树，然后有些点建立了集市。并且没有集市的地方去集市一定是去最近的，如果距离相同，那么则去标号最小的。。现在你还能在建一个集市，问建完这个集市最多有多少个点来这里。。

思路：

现对于每个点求该点到有标记点最近的距离，记录距离及其最近标号，可以用树形dp或者spfa搞。。

然后我们任意选定一个点建树，建完后进行点分治。。

对于当前分治快的跟rt，求rt到每个点的距离为dis，near为到标记点最近的距离

那么对于不同子树的点u,v,如果dis[u] + dis[v] < near[v]，那么如果u建立集市v肯定到u。。

那么我们把式子变形下，dis[u] < near[v] - dis[v],对于u直接二分查找统计即可。。

我们可以先不管哪棵子树先统计一边，然后再减去相同的即可，这样写起来方便点。。写起来还蛮像poj1741的。。

code：

 /*
  * Author:  Yzcstc
  * Created Time:  2014/10/2 17:37:34
  * File Name: hdu5016.cpp
  */
 #pragma comment(linker, "/STACK:102400000,102400000")
 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #include<cstdlib>
 #include<cmath>
 #include<algorithm>
 #include<string>
 #include<map>
 #include<set>
 #include<vector>
 #include<queue>
 #include<stack>
 #include<ctime>
 #define repf(i, a, b) for (int i = (a); i <= (b); ++i)
 #define repd(i, a, b) for (int i = (a); i >= (b); --i)
 #define M0(x)  memset(x, 0, sizeof(x))
 #define clr(x, y) memset(x, y, sizeof(x))
 #define N 120000
 #define Inf 0x3fffffff
 #define x first
 #define y second
 using namespace std;
 typedef pair<int, int> pii;
 struct edge{
      int v, w, next;
      edge(){}
      edge(int _v, int _w, int _next):v(_v), w(_w), next(_next){}
 } e[N<<];
 int last[N], n, len, have[N], ans[N];
 int inq[N];
 pii ner[N];
 
 void add(int u, int v, int w){
     e[len] = edge(v, w, last[u]), last[u] = len++;
 }
 
 void init(){
      clr(last, -);
      len = ;
      int u, v, w;
      for (int i = ; i < n; ++i){
           scanf("%d%d%d", &u, &v, &w);
           add(u, v, w), add(v, u, w);
      }
      repf(i, , n) scanf("%d", &have[i]);
 }
 
 void gao(){
      repf(i, , n) ner[i].x = ner[i].y = Inf, inq[i] = ;
      queue<int> q;
      repf(i, , n) if (have[i]){
            ner[i].x = , ner[i].y = i;
            q.push(i), inq[i] = ;
      }
      int u, v;
      pii tmp;
      while (!q.empty()){
            u = q.front();
            q.pop(), inq[u] = ;
            for (int p = last[u]; ~p; p = e[p].next){
                  v = e[p].v;
                  tmp.x = ner[u].x + e[p].w, tmp.y = ner[u].y;
                  if (tmp < ner[v]){
                         ner[v] = tmp;
                         if (!inq[v]) inq[v] = , q.push(v);
                  }
            }
      }
 
 }
 /*** 点分治-begin **/
 int vis[N], sz, msize[N], f[N], size[N], belong[N], ss, ff[N];
 int dis[N];
 pair<int, int> s[N], tp;
 
 void dfs(int u, int fa){ //calculate the size
      f[sz++] = u;
      msize[u] = , size[u] = ;
      int v;
      for (int p = last[u]; ~p; p = e[p].next){
           v = e[p].v;
           if (v == fa || vis[v]) continue;
           dfs(v, u);
           size[u] += size[v];
           msize[u] = max(size[v], msize[u]);
      }
 }
 
 void dfs(int u, int fa, int dist){
      dis[u] = dist, ff[ss++] = u;
      int v;
      for (int p = last[u]; ~p; p = e[p].next){
           v = e[p].v;
           if (v == fa || vis[v]) continue;
           dfs(v, u, dist + e[p].w);
      }
 }
 
 void calculate(int u, int dist){
      ss = ;
      dfs(u, , dist);
      for (int i = ; i < ss; ++i)
           s[i].x = ner[ff[i]].x - dis[ff[i]], s[i].y = ner[ff[i]].y;
      sort(s, s+ss);
      int t;
      for (int i = ; i < ss; ++i) if (!have[ff[i]]){
           tp.x = dis[ff[i]], tp.y = ff[i];
           t = lower_bound(s, s+ss, tp) - s;
           ans[ff[i]] += (dist > ) ? (t - ss) : (ss - t);
      }
 }
 
 void dfs(int u){
      sz = ;
      dfs(u, );
      int rt = , tmp, v;
      for (int i = ; i < sz; ++i){
            tmp = max(msize[f[i]], sz--size[f[i]]);
            if (tmp <= (sz>>)) { rt = f[i]; break; }
      }
      calculate(rt, );
      vis[rt] = ;
      for (int p = last[rt]; ~p; p = e[p].next){
            v = e[p].v;
            if (vis[v]) continue;
            calculate(v, e[p].w);
            dfs(v);
      }
 }
 /**** 点分治-end***/
 
 void solve(){
      gao();
      M0(vis);
      M0(ans);
      dfs();
      int mx = ;
      for (int i = ; i <= n; ++i)
            mx = max(ans[i], mx);
      printf("%d\n", mx);
 }
 
 int main(){
 //    freopen("a.in", "r", stdin);
 //    freopen("a.out", "w", stdout);
     while (scanf("%d", &n) != EOF){
           init();
           solve();
     }
     return ;
 }