- 1000ms
- 262144K
DSM(Data Structure Master) once learned about tree when he was preparing for NOIP(National Olympiad in Informatics in Provinces) in Senior High School. So when in Data Structure Class in College, he is always absent-minded about what the teacher says.
The experienced and knowledgeable teacher had known about him even before the first class. However, she didn't wish an informatics genius would destroy himself with idleness. After she knew that he was so interested in ACM(ACM International Collegiate Programming Contest), she finally made a plan to teach him to work hard in class, for knowledge is infinite.
This day, the teacher teaches about trees." A tree with nn nodes, can be defined as a graph with only one connected component and no cycle. So it has exactly n-1n−1 edges..." DSM is nearly asleep until he is questioned by teacher. " I have known you are called Data Structure Master in Graph Theory, so here is a problem. "" A tree with nn nodes, which is numbered from 11 to nn. Edge between each two adjacent vertexes uu and vv has a value w, you're asked to answer the number of edge whose value is no more than kk during the path between uu and vv."" If you can't solve the problem during the break, we will call you DaShaMao(Foolish Idiot) later on."
The problem seems quite easy for DSM. However, it can hardly be solved in a break. It's such a disgrace if DSM can't solve the problem. So during the break, he telephones you just for help. Can you save him for his dignity?
Input
In the first line there are two integers n,mn,m, represent the number of vertexes on the tree and queries(2 \le n \le 10^5,1 \le m \le 10^52≤n≤105,1≤m≤105)
The next n-1n−1 lines, each line contains three integers u,v,wu,v,w, indicates there is an undirected edge between nodes uu and vv with value ww. (1 \le u,v \le n,1 \le w \le 10^91≤u,v≤n,1≤w≤109)
The next mm lines, each line contains three integers u,v,ku,v,k , be consistent with the problem given by the teacher above. (1 \le u,v \le n,0 \le k \le 10^9)(1≤u,v≤n,0≤k≤109)
Output
For each query, just print a single line contains the number of edges which meet the condition.
样例输入1
3 3
1 3 2 2 3 7 1 3 0 1 2 4 1 2 7
样例输出1
0
1 2
样例输入2
5 2
1 2 1000000000 1 3 1000000000 2 4 1000000000 3 5 1000000000 2 3 1000000000 4 5 1000000000
样例输出2
2
4
题意简述：给定一棵树，询问m次，求u->v树上路径权值≤k的条数
利用树链剖分+离线线段树进行操作
复杂度为NlogN
1 #include<bits/stdc++.h> 2 3 #define l(x) Tree[x].l 4 #define r(x) Tree[x].r 5 #define sum(x) Tree[x].sum 6 #define ls(x) x << 1 7 #define rs(x) x << 1 | 1 8 9 const int MAXN = (int)1e5 + 5; 10 11 int ver[MAXN << 1], next[MAXN << 1], head[MAXN], tot; 12 int fa[MAXN], son[MAXN], siz[MAXN], dep[MAXN]; 13 int top[MAXN], tid[MAXN], rnk[MAXN], pos; 14 int eid[MAXN]; 15 16 struct segmentT { 17 int l, r; 18 int sum; 19 } Tree[MAXN << 2]; 20 21 void build(int p, int l, int r) { 22 l(p) = l, r(p) = r; 23 if (l == r) return; 24 int mid = (l + r) / 2; 25 build(ls(p), l, mid); 26 build(rs(p), mid + 1, r); 27 } 28 29 void change(int p, int x) { 30 if (l(p) == r(p)) { 31 sum(p) = 1; 32 return; 33 } 34 int mid = (l(p) + r(p)) / 2; 35 if (x <= mid) 36 change(ls(p), x); 37 else 38 change(rs(p), x); 39 sum(p) = sum(ls(p)) + sum(rs(p)); 40 } 41 42 int ask(int p, int l, int r) { 43 if (l <= l(p) && r(p) <= r) return sum(p); 44 int mid = (l(p) + r(p)) / 2; 45 int val = 0; 46 if (l <= mid) val += ask(ls(p), l, r); 47 if (r > mid) val += ask(rs(p), l, r); 48 return val; 49 } 50 51 void add(int u, int v) { 52 ++tot, ver[tot] = v, next[tot] = head[u], head[u] = tot; 53 } 54 55 int dfs1(int u, int f) { 56 dep[u] = dep[f] + 1, siz[u] = 1, son[u] = 0, fa[u] = f; 57 for (int i = head[u]; i; i = next[i]) { 58 int v = ver[i]; 59 if (v == f) continue; 60 siz[u] += dfs1(v, u); 61 eid[(i-1) / 2 + 1] = v; 62 if (siz[v] > siz[son[u]]) son[u] = v; 63 } 64 return siz[u]; 65 } 66 67 void dfs2(int u, int tp) { 68 top[u] = tp, tid[u] = ++pos, rnk[pos] = u; 69 if (!son[u]) return; 70 dfs2(son[u], tp); 71 for (int i = head[u]; i; i = next[i]) { 72 int v = ver[i]; 73 if (v == fa[u] || v == son[u]) continue; 74 dfs2(v, v); 75 } 76 } 77 78 int linkquery(int u, int v) { 79 int ans = 0; 80 while (top[u] != top[v]) { 81 if (dep[top[u]] < dep[top[v]]) std::swap(u, v); 82 ans += ask(1, tid[top[u]], tid[u]); 83 u = fa[top[u]]; 84 } 85 if (u == v) return ans; 86 if (tid[v] < tid[u]) std::swap(u, v); 87 ans += ask(1, tid[u] + 1, tid[v]); 88 return ans; 89 } 90 91 struct node { 92 int u, v, w, id; 93 bool operator<(const node& a) const{ 94 return w < a.w; 95 } 96 } q[MAXN], p[MAXN]; 97 98 int ans[MAXN]; 99 100 int main() { 101 int n, m; 102 scanf("%d%d", &n, &m); 103 for (int i = 1; i < n; i++) { 104 int u, v, w; 105 scanf("%d%d%d", &u, &v, &w); 106 add(u, v), add(v, u); 107 p[i].u = u, p[i].v = v, p[i].w = w, p[i].id = i; 108 } 109 for (int i = 1; i <= m; i++) { 110 int u, v, w; 111 scanf("%d%d%d", &u, &v, &w); 112 q[i].u = u, q[i].v = v, q[i].w = w, q[i].id = i; 113 } 114 std::sort(p + 1, p + n); 115 std::sort(q + 1, q + m + 1); 116 117 dfs1(1, 0); 118 dfs2(1, 1); 119 build(1, 1, n); 120 121 int j; 122 for (int i = 1; i <= m; i++) { 123 while (j < n && p[j].w <= q[i].w) { 124 change(1, tid[eid[p[j].id]]), j++; 125 } 126 ans[q[i].id] = linkquery(q[i].u, q[i].v); 127 } 128 for (int i = 1; i <= m; i++) { 129 printf("%d\n", ans[i]); 130 } 131 return 0; 132 }