牛客练习赛42C（枚举权值）

传送门

思路：既然无法枚举每个情况，那就枚举每个出现过的权值，加和。那么每个权值出现了多少次呢？用总数减去一次都选不中这个数的次数即可，类似概率的方法。

 

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const int maxn = 2001;
const int mod = 1e9 + 7;
int n, m, tot, invn, U, ans;
int id[maxn][maxn], val[maxn * maxn], f[maxn * maxn];
struct node {
    int val, x, y;
    
    bool operator < (const node &rhs) const {
        return val < rhs.val;
    }
}a[maxn * maxn];

inline int ksm(int a, int b) {
    int ret = 1;
    for (; b; b >>= 1) {
        if (b & 1)    ret = (ll)ret * a % mod;
        a = (ll)a * a % mod;
    }
    return ret;
}

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    
    cin >> n >> m;
    invn = ksm(n, mod - 2), U = ksm(n, m);
    
    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <= n; j++) {
            int cid = (i - 1) * n + j;
            cin >> a[cid].val;
            a[cid].x = i, a[cid].y = j;
        }
    }
    sort(a + 1, a + 1 + n * m);
    for (int i = 1; i <= n * m; i++) {
        tot += (i == 1 || a[i - 1].val != a[i].val);
        id[a[i].x][a[i].y] = tot;
        val[tot] = a[i].val;
    }
    
    for (int i = 1; i <= tot; i++)    f[i] = 1;
    for (int i = 1; i <= m; i++) {
        sort(id[i] + 1, id[i] + 1 + n);
        for (int j = 1, pos; j <= n; j = pos + 1) {
            pos = j;
            while (pos < n && id[i][pos] == id[i][pos + 1])    pos++;
            f[id[i][j]] = (ll)f[id[i][j]] * (n - pos + j - 1) % mod * invn % mod;
        }
    }
    
    for (int i = 1; i <= tot; i++) {
        ans = ((ll)ans + (ll)val[i] * (U - (ll)f[i] * U % mod + mod) % mod) % mod;
    }
    cout << ans << endl;
    
    return 0;
}