codeforces 161D 点分治

传送门：https://codeforces.com/problemset/problem/161/D

题意：

求树上点对距离恰好为k的点对个数

题解：

与poj1741相似

把点分治的模板改一下即可，我们依然是求得一个dep数组，然后根据这个dep数组来更新两点间的距离，由于k的范围只有500，所以我们可以直接开一个500的数组来统计两点间距离的数量

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 EDGE {
    int v, w, nxt;
} edge[maxn << 1];
int head[maxn], 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], son[maxn], dep[maxn], vis[maxn];
int Maxt, root, Allnode, cnt;
LL ans;
int n, k;

void get_root(int u, int fa) {
    sz[u] = 1;
    for(int i = head[u]; i != -1; i = edge[i].nxt) {
        int v = edge[i].v;
        if(!vis[v] && v != fa) {
            get_root(v, u);
            sz[u] += sz[v];
        }
    }
    int tmp = max(sz[u] - 1, Allnode - sz[u]);
    if(Maxt > tmp) Maxt = tmp, root = u;
}
void dfs(int u, int fa, int len, int dis) {
    dep[++cnt] = dis;
    if(dis >= len) return;
    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, dis + 1);
        }
    }
}
LL cal(int rt, int fa, int len) {
    if(len <= 0) return len == 0;
    cnt = 0;
    dfs(rt, fa, len, 0);
    LL res = 0;
    int num[505]{};
    for(int i = 1; i <= cnt; i++) {
        num[dep[i]]++;
    }
    for(int i = 1; i <= cnt; i++) {
        res += num[len - dep[i]];
    }
    return res;
}
void divide(int rt) {
    vis[rt] = 1;
    // debug1(ans);
    ans += cal(rt, 0, k);
    for(int i = head[rt]; i != -1; i = edge[i].nxt) {
        int v = edge[i].v;
        if(!vis[v]) {
            ans -= cal(v, rt, k - 2);
            Allnode = sz[v];
            Maxt = n;
            get_root(v, rt);
            divide(root);
        }
    }
}
int main() {
#ifndef ONLINE_JUDGE
    FIN
#endif
    while(scanf("%d%d", &n, &k) != EOF) {
        memset(head, -1, sizeof(head));
        tot = 0;
        for(int i = 1, u, v; i < n; i++) {
            scanf("%d%d", &u, &v);
            add_edge(u, v, 1);
            add_edge(v, u, 1);
        }
        memset(vis, 0, sizeof(vis));
        Allnode = n;
        Maxt = INF;
        get_root(1, 0);
        divide(root);
        printf("%lld
", ans/2);
    }
    return 0;
}