2020 CCPC Changchun F :Strange Memory dsu on tree-优快云博客

本文链接：https://blog.youkuaiyun.com/tlyzxc/article/details/121071879

这篇博客讨论了一种在树形结构中进行静态统计的方法，利用dfsu（深度优先搜索并更新）来维护每个节点的二进制位1出现的次数，从而能快速计算异或值。代码实现中，通过维护一个二维数组nums记录每个数值每个二进制位的1出现次数，并在dsu过程中更新这些信息，最后计算异或和。

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分析

一开始读错题了，以为是个算贡献的水题。。。
树上静态统计，考虑 $\ on \ tree$ ，我们去维护每一个值对应每一个二进制位出现了多少次1，然后就可以快速计算异或值

代码

#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define debug(x) cout<<#x<<":"<<x<<endl;
#define dl(x) printf("%lld\n",x);
#define di(x) printf("%d\n",x);
#define _CRT_SECURE_NO_WARNINGS
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef vector<ll> VI;
const int INF = 0x3f3f3f3f;
const int N = 2e5 + 10, M = N * 2;
const ll mod = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
template<typename T>inline void read(T &a) {
    char c = getchar(); T x = 0, f = 1; while (!isdigit(c)) {if (c == '-')f = -1; c = getchar();}
    while (isdigit(c)) {x = (x << 1) + (x << 3) + c - '0'; c = getchar();} a = f * x;
}
int gcd(int a, int b) {return (b > 0) ? gcd(b, a % b) : a;}
int h[N], e[M], ne[M], idx;
int sz[N], son[N], cnt[N];
int nums[1000010][25][2];
int n, Son, mx;
ll sum,ans;
int a[N];

void add(int x, int y) {
    ne[idx] = h[x], e[idx] = y, h[x] = idx++;
}

void dfs(int u, int fa) {
    sz[u] = 1;
    for (int i = h[u]; ~i; i = ne[i]) {
        int j = e[i];
        if (j ==  fa) continue;
        dfs(j, u);
        sz[u] += sz[j];
        if (sz[son[u]] < sz[j]) son[u] = j;
    }
}

void cal(int u, int fa, int st) {
    for(int i = 24;~i;i--){
        nums[a[u]][i][(u >> i) & 1] += st;
    }
    for (int i = h[u]; ~i; i = ne[i]) {
            if (e[i] != Son && e[i] != fa)
                cal(e[i], u, st);
    }
}

void count(int u,int fa,int y){
    int val = y ^ a[u];
    if(val < 1000010){
        for(int i = 24;~i;i--){
            int p = (u >> i) & 1;
            ans += (1 << i) * nums[val][i][p ^ 1];
        }
    }
    for (int i = h[u]; ~i; i = ne[i]) {
        if (e[i] != Son && e[i] != fa)
            count(e[i],u,y);
    }
}

void dsu(int u, int fa, bool st) {
    for (int i = h[u]; ~i; i = ne[i]) {
        int j = e[i];
        if (j == fa || j == son[u]) continue;
        dsu(j, u, 0);
    }
    if (son[u]) dsu(son[u], u, 1);
    Son = son[u];
    for(int i = h[u];~i;i = ne[i]){
        int j = e[i];
        if(j == fa || j == son[u]) continue;
        count(j,u,a[u]);
        cal(j,u,1);
    }
    int val = a[u];
    for(int i = 24;~i;i--){
        nums[val][i][(u >> i) & 1] += 1;
    }
    Son = 0;
    if (!st) cal(u, fa, -1);
}

int main() {
    memset(h, -1, sizeof h);
    read(n);
    for (int i = 1; i <= n; i++) read(a[i]);
    for (int i = 1; i < n; i++) {
        int a, b;
        read(a), read(b);
        add(a, b), add(b, a);
    }
    dfs(1, 0);
    dsu(1, 0, 0);
    dl(ans);
    return 0;
}