BZOJ 1316: 树上的询问

挺裸的点分治

刚开始想用map水过去，然后做p次点分治，然后T到自闭

最后发现可以sort一遍，然后去重，记录每个数出现的次数，这样就可以双指针，不会漏掉了

 #include <bits/stdc++.h>
 using namespace std;

 #define N 100010
 int n, q, Q[N], ans[N];
 struct Graph
 {
     struct node
     {
         int to, nx, w;
         node() {}
         node (int to, int nx, int w) : to(to), nx(nx), w(w) {}
     }a[N << ];
     int head[N], pos;
     void add(int u, int v, int w)
     {
         a[++pos] = node(v, head[u], w); head[u] = pos;
         a[++pos] = node(u, head[v], w); head[v] = pos;
     }
 }G;
 #define erp(u) for (int it = G.head[u], v = G.a[it].to, w = G.a[it].w; it; it = G.a[it].nx, v = G.a[it].to, w = G.a[it].w)

 bool vis[N];
 int root, sum, sze[N], f[N];
 void getroot(int u, int fa)
 {
     sze[u] = , f[u] = ;
     erp(u) if (v != fa && !vis[v])
     {
         getroot(v, u);
         sze[u] += sze[v];
         f[u] = max(f[u], sze[v]);
     }
     f[u] = max(f[u], sum - sze[u]);
     if (f[u] < f[root]) root = u;
 }

 int deep[N], d[N];
 void getdeep(int u, int fa)
 {
     deep[++deep[]] = d[u];
     erp(u) if (v != fa && !vis[v])
     {
         d[v] = d[u] + w;
         getdeep(v, u);
     }
 }

 int g[N], num[];
 void calc(int u, int cost, int opt)
 {
     d[u] = cost; deep[] = ;
     getdeep(u, );
     sort(deep + , deep +  + deep[]);
     g[] = ;
     for (int i = ; i <= deep[]; ++i)
     {
         if (i ==  || deep[i] != deep[i - ]) g[++g[]] = deep[i], num[g[]] = ;
         else ++num[g[]];
     }
     for (int i = ; i <= q; ++i)
     {
         for (int j = ; j <= g[]; ++j)
         {
             if (num[j] >  &&  * g[j] == Q[i])
                 ans[i] += opt * num[j] * (num[j] - );
         }
         int r = g[];
         for (int l = ; l < r; ++l)
         {
             while (l < r && g[l] + g[r] >= Q[i])
             {
                 if (g[l] + g[r] == Q[i]) ans[i] += opt * num[l] * num[r];
                 --r;
             }
         }
     }
 }

 void solve(int u)
 {
     calc(u, , );
     vis[u] = ;
     erp(u) if (!vis[v])
     {
         calc(v, w, -);
         sum = f[] = sze[v];
         root = ;
         getroot(v, u);
         solve(root);
     }
 }

 void Run()
 {
     scanf("%d%d", &n, &q);
     {
         for (int i = , u, v, w; i < n; ++i)
         {
             scanf("%d%d%d", &u, &v, &w);
             G.add(u, v, w);
         }
         for (int qq = ; qq <= q; ++qq) scanf("%d", Q + qq);
         sum = n; f[] = n;
         root = ;
         getroot(, );
         solve(root);
         for (int i = ; i <= q; ++i) puts(ans[i] || !Q[i] ? "Yes" : "No");
     }
 }

 int main()
 {
     Run();
 }