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

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3  
  4 #define N 100010
  5 int n, q, Q[N], ans[N]; 
  6 struct Graph
  7 {
  8     struct node
  9     {
 10         int to, nx, w;
 11         node() {}
 12         node (int to, int nx, int w) : to(to), nx(nx), w(w) {}
 13     }a[N << 1];
 14     int head[N], pos;
 15     void add(int u, int v, int w)
 16     {
 17         a[++pos] = node(v, head[u], w); head[u] = pos; 
 18         a[++pos] = node(u, head[v], w); head[v] = pos; 
 19     }
 20 }G;
 21 #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)
 22  
 23 bool vis[N];
 24 int root, sum, sze[N], f[N]; 
 25 void getroot(int u, int fa)
 26 {
 27     sze[u] = 1, f[u] = 0;
 28     erp(u) if (v != fa && !vis[v])
 29     {
 30         getroot(v, u);
 31         sze[u] += sze[v];
 32         f[u] = max(f[u], sze[v]);   
 33     }
 34     f[u] = max(f[u], sum - sze[u]);
 35     if (f[u] < f[root]) root = u; 
 36 }
 37  
 38 int deep[N], d[N];
 39 void getdeep(int u, int fa)
 40 {
 41     deep[++deep[0]] = d[u];
 42     erp(u) if (v != fa && !vis[v])
 43     {
 44         d[v] = d[u] + w; 
 45         getdeep(v, u);
 46     }
 47 }
 48  
 49 int g[N], num[10000010]; 
 50 void calc(int u, int cost, int opt)
 51 {
 52     d[u] = cost; deep[0] = 0;
 53     getdeep(u, 0);
 54     sort(deep + 1, deep + 1 + deep[0]);
 55     g[0] = 0;
 56     for (int i = 1; i <= deep[0]; ++i)
 57     {
 58         if (i == 1 || deep[i] != deep[i - 1]) g[++g[0]] = deep[i], num[g[0]] = 1;
 59         else ++num[g[0]]; 
 60     }
 61     for (int i = 1; i <= q; ++i)
 62     {
 63         for (int j = 1; j <= g[0]; ++j) 
 64         {
 65             if (num[j] > 1 && 2 * g[j] == Q[i]) 
 66                 ans[i] += opt * num[j] * (num[j] - 1);
 67         }
 68         int r = g[0];
 69         for (int l = 1; l < r; ++l)
 70         {
 71             while (l < r && g[l] + g[r] >= Q[i]) 
 72             {
 73                 if (g[l] + g[r] == Q[i]) ans[i] += opt * num[l] * num[r];
 74                 --r;
 75             }
 76         }
 77     }
 78 }
 79  
 80 void solve(int u)
 81 {
 82     calc(u, 0, 1);
 83     vis[u] = 1;
 84     erp(u) if (!vis[v]) 
 85     {
 86         calc(v, w, -1);
 87         sum = f[0] = sze[v]; 
 88         root = 0;
 89         getroot(v, u); 
 90         solve(root); 
 91     }
 92 }
 93  
 94 void Run()
 95 {
 96     scanf("%d%d", &n, &q);
 97     {
 98         for (int i = 1, u, v, w; i < n; ++i)
 99         {
100             scanf("%d%d%d", &u, &v, &w);
101             G.add(u, v, w);
102         }
103         for (int qq = 1; qq <= q; ++qq) scanf("%d", Q + qq);
104         sum = n; f[0] = n;
105         root = 0; 
106         getroot(1, 0);
107         solve(root);
108         for (int i = 1; i <= q; ++i) puts(ans[i] || !Q[i] ? "Yes" : "No");
109     }
110 }
111  
112 int main()
113 {
114     Run();
115 }
View Code

 

posted @ 2018-12-04 18:38  Dup4  阅读(156)  评论(0)  编辑  收藏  举报