主要思路
这道题应该算是点分治的一道经典题目。点分治的精髓就在于找到重心、对子树进行计算并且合并答案，之后删除中心变成子树内的小问题，分治思想非常的巧妙。
我们首先写好找根函数：
// root finding functions;
void dfs1(int u, int fa, int siz)
{
son[u] = 1;
int res = 0;
for (int i = head[u]; i != -1; i = edges[i].nxt)
{
if (edges[i].to == fa || done[edges[i].to])
continue;
dfs1(edges[i].to, u, siz);
son[u] += son[edges[i].to];
res = max(res, son[edges[i].to] - 1);
}
res = max(res, siz - son[u]);
if (res < rootKey)
root = u, rootKey = res;
}
void findRoot(int src, int siz)
{
root = src, rootKey = siz;
dfs1(src, 0, siz);
}
遍历子树：我们设定一个\(dist[]\)数组，算出距离的答案，并且无视曾经被当作根的节点，并向cnt[from[u]]进行自增操作，维护子树的大小，并把本身编号加入vec供之后进行排序用途。
// the calc one;
void dfs(int u, int fa)
{
vec.push_back(u);
for (int i = head[u]; i != -1; i = edges[i].nxt)
if (edges[i].to != fa && !done[edges[i].to])
{
cnt[from[u]]++;
dist[edges[i].to] = dist[u] + edges[i].weight;
from[edges[i].to] = from[u];
dfs(edges[i].to, u);
}
}
点分治：我们在计算完子树答案之后，合并的方法是主要的问题。我们可以考虑将vec进行排序，然后通过递增的性质设置首尾指针，并且去除掉当前子树内的错误答案（因为子树内部的路径不经过根，所以要去掉，之后的分治子问题中会处理这些内部路径）。
void pdq(int curt, int siz)
{
memset(son, 0, sizeof(son));
memset(cnt, 0, sizeof(cnt));
findRoot(curt, siz);
dist[root] = 0, done[root] = true;
vec.clear(), vec.push_back(root), from[root] = root;
cnt[root]++;
for (int i = head[root]; i != -1; i = edges[i].nxt)
if (!done[edges[i].to])
{
from[edges[i].to] = edges[i].to, cnt[edges[i].to]++;
dist[edges[i].to] = edges[i].weight;
dfs(edges[i].to, root);
}
sort(vec.begin(), vec.end(), compare);
cnt[from[vec[0]]]--;
int l = 0, r = vec.size() - 1;
while (l < r)
{
while (dist[vec[l]] + dist[vec[r]] > k)
cnt[from[vec[r--]]]--;
answer += r - l - cnt[from[vec[l]]];
cnt[from[vec[++l]]]--;
}
int pos = root;
for (int i = head[pos]; i != -1; i = edges[i].nxt)
if (!done[edges[i].to])
pdq(edges[i].to, son[edges[i].to]);
}
嗯，写完了。具体代码如下：
代码
// POJ1741.cpp
#include <cstdio>
#include <vector>
#include <algorithm>
#include <iostream>
#include <cstring>
using namespace std;
const int MAX_N = 10100, INF = 0x3f3f3f3f;
int head[MAX_N], current, k, n, tmpx, tmpy, tmpz, root, rootKey, son[MAX_N];
int from[MAX_N], dist[MAX_N], cnt[MAX_N];
bool done[MAX_N];
long long answer = 0;
vector<int> vec;
struct edge
{
int to, nxt, weight;
} edges[MAX_N << 1];
bool compare(int a, int b) { return dist[a] < dist[b]; }
void addpath(int src, int dst, int weight)
{
edges[current].to = dst, edges[current].nxt = head[src];
edges[current].weight = weight, head[src] = current++;
}
// root finding functions;
void dfs1(int u, int fa, int siz)
{
son[u] = 1;
int res = 0;
for (int i = head[u]; i != -1; i = edges[i].nxt)
{
if (edges[i].to == fa || done[edges[i].to])
continue;
dfs1(edges[i].to, u, siz);
son[u] += son[edges[i].to];
res = max(res, son[edges[i].to] - 1);
}
res = max(res, siz - son[u]);
if (res < rootKey)
root = u, rootKey = res;
}
void findRoot(int src, int siz)
{
root = src, rootKey = siz;
dfs1(src, 0, siz);
}
// the calc one;
void dfs(int u, int fa)
{
vec.push_back(u);
for (int i = head[u]; i != -1; i = edges[i].nxt)
if (edges[i].to != fa && !done[edges[i].to])
{
cnt[from[u]]++;
dist[edges[i].to] = dist[u] + edges[i].weight;
from[edges[i].to] = from[u];
dfs(edges[i].to, u);
}
}
void pdq(int curt, int siz)
{
memset(son, 0, sizeof(son));
memset(cnt, 0, sizeof(cnt));
findRoot(curt, siz);
dist[root] = 0, done[root] = true;
vec.clear(), vec.push_back(root), from[root] = root;
cnt[root]++;
for (int i = head[root]; i != -1; i = edges[i].nxt)
if (!done[edges[i].to])
{
from[edges[i].to] = edges[i].to, cnt[edges[i].to]++;
dist[edges[i].to] = edges[i].weight;
dfs(edges[i].to, root);
}
sort(vec.begin(), vec.end(), compare);
cnt[from[vec[0]]]--;
int l = 0, r = vec.size() - 1;
while (l < r)
{
while (dist[vec[l]] + dist[vec[r]] > k)
cnt[from[vec[r--]]]--;
answer += r - l - cnt[from[vec[l]]];
cnt[from[vec[++l]]]--;
}
int pos = root;
for (int i = head[pos]; i != -1; i = edges[i].nxt)
if (!done[edges[i].to])
pdq(edges[i].to, son[edges[i].to]);
}
int main()
{
while (scanf("%d%d", &n, &k) && n != 0)
{
answer = 0;
memset(head, -1, sizeof(head)), current = 0;
for (int i = 1; i <= n - 1; i++)
scanf("%d%d%d", &tmpx, &tmpy, &tmpz), addpath(tmpx, tmpy, tmpz), addpath(tmpy, tmpx, tmpz);
memset(done, false, sizeof(done));
pdq(1, n);
printf("%lld\n", answer);
}
return 0;
}
