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 = ;
 const int mod = 1e9 + ;
 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 = ;
     for (; b; b >>= ) {
         if (b & )    ret = (ll)ret * a % mod;
         a = (ll)a * a % mod;
     }
     return ret;
 }

 int main() {
     ios_base::sync_with_stdio();
     cin.tie();

     cin >> n >> m;
     invn = ksm(n, mod - ), U = ksm(n, m);

     for (int i = ; i <= m; i++) {
         for (int j = ; j <= n; j++) {
             int cid = (i - ) * n + j;
             cin >> a[cid].val;
             a[cid].x = i, a[cid].y = j;
         }
     }
     sort(a + , a +  + n * m);
     for (int i = ; i <= n * m; i++) {
         tot += (i ==  || a[i - ].val != a[i].val);
         id[a[i].x][a[i].y] = tot;
         val[tot] = a[i].val;
     }

     for (int i = ; i <= tot; i++)    f[i] = ;
     for (int i = ; i <= m; i++) {
         sort(id[i] + , id[i] +  + n);
         for (int j = , pos; j <= n; j = pos + ) {
             pos = j;
             while (pos < n && id[i][pos] == id[i][pos + ])    pos++;
             f[id[i][j]] = (ll)f[id[i][j]] * (n - pos + j - ) % mod * invn % mod;
         }
     }

     for (int i = ; i <= tot; i++) {
         ans = ((ll)ans + (ll)val[i] * (U - (ll)f[i] * U % mod + mod) % mod) % mod;
     }
     cout << ans << endl;

     return ;
 }