Codeforces 293E
传送门：https://codeforces.com/contest/293/problem/E
题意：
给你一颗边权一开始为0的树，然后给你n-1次操作，每次给边加上边权，问你n-1次操作后有有多少对点之间的路径长度小于等于l，并且边权和小于等于w
题解：
poj1741 点分治裸题是 边权和小于等于k，这里加了一个路径条数的限制
对于这个路径条数和边权的两个限制，我们可以得到两个不等式，可以用点分治得到满足距离的一个数组a
将数组a按照距离从小到大排序后，就可以满足cdq的条件了，也是一个三维偏序问题，计数的时候需要注意一下去重
具体解释请看代码注释
代码：
#include <set>
#include <map>
#include <cmath>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
#define ls rt<<1
#define rs rt<<1|1
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define bug printf("*********
")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]
"
#define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]
"
#define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]
"
const int maxn = 3e5 + 5;
const int INF = 0x3f3f3f3f;
struct node {
int x, y;
int op;
bool operator < (const node &a) const {
if(x != a.x) return x < a.x;
if(y != a.y) return y < a.y;
return op < a.op;
}
node() {};
node(int _x, int _y, int _op) {
x = _x, y = _y, op = _op;
}
} a[maxn], tmp[maxn];
int cnt;
int n, L, W;
struct EDGE {
int v, w, nxt;
} edge[maxn << 1];
int head[maxn];
int tot;
void add_edge(int u, int v, int w) {
edge[tot].v = v;
edge[tot].w = w;
edge[tot].nxt = head[u];
head[u] = tot++;
}
int sz[maxn], vis[maxn], mx[maxn];
int get_root(int u, int fa, int num) {
int y = 0;
mx[u] = 0;
sz[u] = 1;
for(int i = head[u]; i != -1; i = edge[i].nxt) {
int v = edge[i].v;
if(!vis[v] && v != fa) {
int z = get_root(v, u, num);
// debug3(u, v, z);
sz[u] += sz[v];
mx[u] = max(mx[u], sz[v]);
if(mx[y] > mx[z]) y = z;
}
}
mx[u] = max(mx[u], num - sz[u]);
return mx[u] < mx[y] ? u : y;
}
void dfs(int u, int fa, int len, int weight) {
a[++cnt] = node(len, weight, 0);//与根节点的距离和权值 当前点 op为0
a[++cnt] = node(L - len, W - weight, 1);//还剩下可以走的距离与权值限制 限制 op为1
for(int i = head[u]; i != -1; i = edge[i].nxt) {
int v = edge[i].v;
if(!vis[v] && v != fa) {
dfs(v, u, len + 1, weight + edge[i].w);
}
}
}
LL cdq(int l, int r) {
if(l == r) return 0;
int mid = (l + r) >> 1;
LL ans = cdq(l, mid) + cdq(mid + 1, r);
int p = l, q = mid + 1, res = 0;
for(int i = l; i <= r; i++) {//因为左边的x已经小于右边的x了，所以只要比较y就行
if((p <= mid && (a[p].y < a[q].y || (a[p].y == a[q].y && a[p].op <= a[q].op))) || q > r) {
res += a[p].op ^ 1;//如果这个点是非限制点，res++
tmp[i] = a[p++];//恢复现场
} else {
ans += a[q].op * res;//如果这个点是限制点,答案就可以增加
tmp[i] = a[q++];
}
}
for(int i = l; i <= r; i++) {
a[i] = tmp[i];
}
// debug3(l, r, ans);
return ans;
}
LL Find(int u, int len, int weight) {
LL res = 0; cnt = 0;
dfs(u, -1, len, weight);//获取a数组，即满足l，w限制的数组
// debug1(cnt);
sort(a + 1, a + cnt + 1);//对x排序
for(int i = 1; i <= cnt; i++) {
if(2 * a[i].x <= L && 2 * a[i].y <= W)
res += a[i].op ^ 1;//不满足条件的，op为0就不满足，
}
// debug1(cnt);
debug1(res);
return (cdq(1, cnt) - res) / 2;//a，b，b，a是一样的，所以除二
}
LL solve(int u, int num) {
int root = get_root(u, -1, num);//点分治 找重心
debug1(root);
cout<<"1"<<endl;
LL res = Find(root, 0, 0);//以该重心分治的贡献
cout<<"root"<<root<<"de gong xian is: ";
debug1(res);
vis[root] = 1;
for(int i = head[root]; i != -1; i = edge[i].nxt) {
int v = edge[i].v, w = edge[i].w;
if(!vis[v]) {
cout<<"2"<<endl;
res -= Find(v, 1, w);//因为这个点是由父亲节点跑过来的，所以边长为1的点重复算了需要减去
// debug1(res);//减去重复的
res += solve(v, sz[v] > sz[root] ? num - sz[root] : sz[v]);//子树大小的判断
}
}
return res;
}
int main() {
#ifndef ONLINE_JUDGE
FIN
#endif
mx[0] = INF;
memset(head, -1, sizeof(head));
tot = 0;
scanf("%d%d%d", &n, &L, &W);
for(int i = 2; i <= n; i++) {
int u, w; scanf("%d%d", &u, &w);
add_edge(i, u, w);
add_edge(u, i, w);
}
printf("%lld
", solve(1, n));
return 0;
}