HDU6411 带劲的and和

并查集 + 位运输

先用并查集处理每个联通块，并把权值排序。

然后对于 a * (c & d) 可以把c和d才成二进制，同样位置上存在1才对答案有贡献，所以我们对每个联通块，枚举每个位置是1的数量，算出贡献就好了。

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
#define FAST_IO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
using namespace std;
typedef long long ll;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int ret = 0, w = 0; char ch = 0;
    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
    while(isdigit(ch)) ret = (ret << 3) + (ret << 1) + (ch ^ 48), ch = getchar();
    return w ? -ret : ret;
}
inline int gcd(int a, int b){ return b ? gcd(b, a % b) : a; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
template <typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template <typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template <typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
    return ans;
}
const int N = 100005;
const int MOD = 1e9 + 7;
int _, n, m, v[N], parent[N], t[30];
vector<int> vi[N];
int find(int p){
    while(p != parent[p]){
        parent[p] = parent[parent[p]];
        p = parent[p];
    }
    return p;
}

bool isConnect(int p, int q){
    return find(p) == find(q);
}

void merge(int p, int q){
    int pRoot = find(p), qRoot = find(q);
    if(pRoot == qRoot) return;
    parent[qRoot] = pRoot;
}

void build(){
    for(int i = 0; i <= n; i ++){
        parent[i] = i;
        vi[i].clear();
    }
}

int main(){

    for(_ = read(); _; _ --){
        n = read(), m = read(), build();
        for(int i = 1; i <= n; i ++) v[i] = read();
        for(int i = 1; i <= m; i ++){
            int x = read(), y = read();
            if(isConnect(x, y)) continue;
            merge(x, y);
        }
        for(int i = 1; i <= n; i ++) vi[find(i)].push_back(v[i]);
        for(int i = 1; i <= n; i ++) sort(vi[i].begin(), vi[i].end());
        ll ans = 0;
        for(int i = 1; i <= n; i ++){
            if(vi[i].size() <= 1) continue;
            full(t, 0);
            for(auto val: vi[i]){
                for(int j = 0; j < 32; j ++){
                    if(val & (1 << j)){
                        ans = (ans % MOD + 1LL * t[j] * val % MOD * (1 << j) % MOD) % MOD;
                        t[j] ++;
                    }
                }
            }
        }
        printf("%lld
", ans);
    }
    return 0;
}