CodeForces - 766E Mahmoud and a xor trip

题意：

给定一棵有 $n$ 个结点的树，点带权，求所有路径异或值之和。$(n\le 10^5)$

链接：

https://codeforces.com/problemset/problem/766/E

解题思路：

若是路径权值和，则树形 $dp$ 就可，现在变成异或，按位考虑。异或值之和可拆位表示成各二进制位数量，树形 $dp$ 时维护子树内各二进制位数量，则可以转移维护。

参考代码：

#include
 
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
#define sz(a) ((int)a.size())
#define pb push_back
#define lson (rt << 1)
#define rson (rt << 1 | 1)
#define gmid (l + r >> 1)
const int maxn = 1e5 + 5;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
 
vector<int> G[maxn];
int a[maxn], dp[maxn][20][2];
int n; ll ans;

void dfs(int u, int f){

    for(int i = 0; i < 20; ++i){

        ++dp[u][i][(a[u] >> i) & 1];
    }
    for(auto &v : G[u]){

        if(v == f) continue;
        dfs(v, u);
        for(int i = 0; i < 20; ++i){

            ans += (1ll << i) * dp[v][i][0] * dp[u][i][1];
            ans += (1ll << i) * dp[v][i][1] * dp[u][i][0];
        }
        for(int i = 0; i < 20; ++i){

            int t = (a[u] >> i) & 1;
            dp[u][i][0] += dp[v][i][t];
            dp[u][i][1] += dp[v][i][t ^ 1];
        }
    }
}


int main(){
 
    ios::sync_with_stdio(0); cin.tie(0);
    cin >> n;
    for(int i = 1; i <= n; ++i) cin >> a[i];
    for(int i = 1; i < n; ++i){
        
        int u, v; cin >> u >> v;
        G[u].pb(v), G[v].pb(u);
    }
    dfs(1, 0);
    for(int i = 1; i <= n; ++i) ans += a[i];
    cout << ans << endl;
    return 0;
}

 
这是为了练习点分 （脑抽） 写的：

#include
 
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
#define sz(a) ((int)a.size())
#define pb push_back
#define lson (rt << 1)
#define rson (rt << 1 | 1)
#define gmid (l + r >> 1)
const int maxn = 1e5 + 5;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
 
vector<int> G[maxn];
int a[maxn], siz[maxn], vis[maxn], dis[maxn], mp[20][2];
int n, tn, rmn, rt, tot;
ll ans;
 
void getRt(int u, int f){
 
    int mx = 0; siz[u] = 1; 
    for(auto &v : G[u]){
 
        if(v == f || vis[v]) continue;
        getRt(v, u);
        siz[u] += siz[v];
        mx = max(mx, siz[v]);
    }
    mx = max(mx, tn - siz[u]);
    if(mx < rmn) rmn = mx, rt = u;
}
 
void add(int x, int val){
 
    for(int i = 0; i < 20; ++i){
 
        mp[i][(x >> i) & 1] += val;
    }
}
 
void dfs(int u, int f, int val){
 
    dis[++tot] = val;
    for(auto &v : G[u]){
 
        if(v == f || vis[v]) continue;
        dfs(v, u, val ^ a[v]);
    }
}
 
void cal(int u){
 
    add(0, 1);
    for(auto &v : G[u]){
 
        if(vis[v]) continue;
        tot = 0;
        dfs(v, u, a[v]);
        for(int i = 1; i <= tot; ++i){
 
            int tmp = dis[i] ^ a[u];
            for(int j = 0; j < 20; ++j){
 
                ans += (1ll << j) * mp[j][(~tmp >> j) & 1];
            }
        }
        for(int i = 1; i <= tot; ++i){
 
            add(dis[i], 1);
        }
    }
    memset(mp, 0, sizeof mp);
}
 
void dfz(int u){
 
    vis[u] = 1;
    cal(u);
    for(auto &v : G[u]){
 
        if(vis[v]) continue;
        tn = siz[v], rmn = inf, getRt(v, u);
        dfz(rt);
    }
    vis[u] = 0;
}
 
int main(){
 
    ios::sync_with_stdio(0); cin.tie(0);
    cin >> n;
    for(int i = 1; i <= n; ++i) cin >> a[i];
    for(int i = 1; i < n; ++i){
        
        int u, v; cin >> u >> v;
        G[u].pb(v), G[v].pb(u);
    }
    tn = n, rmn = inf, getRt(1, 0);
    dfz(rt);
    for(int i = 1; i <= n; ++i) ans += a[i];
    cout << ans << endl;
    return 0;
}

 
这是为了练习换根 $dp$ （脑抽） 写的：

#include
 
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
#define sz(a) ((int)a.size())
#define pb push_back
#define lson (rt << 1)
#define rson (rt << 1 | 1)
#define gmid (l + r >> 1)
const int maxn = 1e5 + 5;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
 
vector<int> G[maxn];
int a[maxn], dp[maxn][20][2], fp[maxn][20][2];
int n; ll ans;
 
void dfs1(int u, int f){
 
    for(int i = 0; i < 20; ++i){
 
        ++dp[u][i][(a[u] >> i) & 1];
    }
    for(auto &v : G[u]){
 
        if(v == f) continue;
        dfs1(v, u);
        for(int i = 0; i < 20; ++i){
 
            int t = (a[u] >> i) & 1;
            dp[u][i][0] += dp[v][i][t];
            dp[u][i][1] += dp[v][i][t ^ 1];
        }
    }
}
 
void dfs2(int u, int f){
 
    for(auto &v : G[u]){
 
        if(v == f) continue;
        for(int i = 0; i < 20; ++i){
 
            int tu = (a[u] >> i) & 1, tv = (a[v] >> i) & 1;
            fp[v][i][0] += fp[u][i][tv];
            fp[v][i][1] += fp[u][i][tv ^ 1];
            int tp[2] = {};
            tp[0] = dp[u][i][0] - dp[v][i][tu] - (tu ^ 1);
            tp[1] = dp[u][i][1] - dp[v][i][tu ^ 1] - tu;
            fp[v][i][0] += tp[tv] + (tv ^ 1);
            fp[v][i][1] += tp[tv ^ 1] + tv;
        }
        dfs2(v, u);
    }
} 
 
int main(){
 
    ios::sync_with_stdio(0); cin.tie(0);
    cin >> n;
    for(int i = 1; i <= n; ++i) cin >> a[i];
    for(int i = 1; i < n; ++i){
        
        int u, v; cin >> u >> v;
        G[u].pb(v), G[v].pb(u);
    }
    dfs1(1, 0);
    for(int i = 0; i < 20; ++i){
 
        ++fp[1][i][(a[1] >> i) & 1];
    }
    dfs2(1, 0);
    for(int i = 1; i <= n; ++i){
 
        for(int j = 0; j < 20; ++j){
 
            ans += (1ll << j) * (dp[i][j][1] + fp[i][j][1]);
        }
    }
    ans >>= 1;
    cout << ans << endl;
    return 0;
}