題意：給出一個n*m的矩陣，每個點有一種顏色，定義矩陣的val為矩陣中不同顏色的數量，問任意一個子矩陣的val的期望為多少。

思路：我是直接看了一篇講解的想不理解都難的部落格：點選開啟連結

程式碼：

#include<bits/stdc++.h>
#define ll long long
#define pb push_back
#define fi first
#define se second
#define pi acos(-1)
#define inf 0x3f3f3f3f
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define rep(i,x,n) for(int i=x;i<n;i++)
#define per(i,n,x) for(int i=n;i>=x;i--)
using namespace std;
typedef pair<int,int>P;
const int MAXN=10010;
int gcd(int a,int b){return b?gcd(b,a%b):a;}
int n, m;
vector<P> color[MAXN];
vector<int> Y[110];
int last[110];
ll calc(int col)
{
	ll ans = 0;
	memset(last, 0, sizeof(last));
	for(int o = 0; o < color[col].size(); o++)
	{
		int tx = color[col][o].fi, ty = color[col][o].se;
		for(int i = 1; i <= m; i++)
		Y[i].clear();
		for(int i = 1; i <= m; i++)
		if(last[i])
		Y[last[i]].pb(i);
		int tl = 1, tr = m;
		bool flag = 0;
		for(int i = tx; i >= 1; i--)
		{
			for(int j = 0; j < Y[i].size(); j++)
			{
				int tmp = Y[i][j];
				if(tmp < ty)
				tl = max(tl, tmp + 1);
				else if(tmp > ty)
				tr = min(tr, tmp - 1);
				else
				{
					flag = 1;
					break;
				}
			}
			if(flag) break;
			ans += 1ll * (n - tx + 1) * (ty - tl + 1) * (tr - ty + 1); 
		}
		last[ty] = tx;
	}
	return ans;
}
int main()
{
	int T, c;
	scanf("%d", &T);
	while(T--)
	{
		scanf("%d %d", &n, &m);
		for(int i = 0; i <= n * m; i++)
		color[i].clear();
		for(int i = 1; i <= n; i++)
		for(int j = 1; j <= m; j++)
		scanf("%d", &c), color[c].pb(P(i, j));
		for(int i = 0; i <= n * m; i++)
		if(color[i].size())
		sort(color[i].begin(), color[i].end());
		ll sum = 0, ans = 0;
		for(int i = 1; i <= n; i++)
		for(int j = 1; j <= m; j++)
		sum += 1ll * i * j;
		for(int i = 0; i <= n * m; i++)
		if(color[i].size())
		ans += calc(i);
		printf("%.9lf\n", (double)ans * 1.0 / sum);
	}
 	return 0;
}