题意略。
思路:我们单独考虑每种颜色的贡献,颜色c对答案的贡献 == 含有c的矩形个数,这就是在考查我们计数的能力,暴力可过。
#include<bits/stdc++.h> #define maxn 105 using namespace std; int color[maxn][maxn]; int n,m; double calcu(int x,int y){ int l = 0,r = m + 1,c = color[x][y]; double ret = 0; for(int i = x;i >= 1;--i){ if(color[i][y] == c && i < x) break; for(int j = y - 1;j >= 1;--j){ if(color[i][j] == c){ l = max(l,j); break; } } if(i == x){ ret += max(1,(y - l)) * (m - y + 1) * (n - x + 1); continue; } for(int j = y + 1;j <= m;++j){ if(color[i][j] == c){ r = min(r,j); break; } } //printf("l == %d r == %d ",l,r); ret += max(1,(y - l)) * max(1,(r - y)) * (n - x + 1); } //printf("now ret == %lf ",ret); return ret; } int main(){ int T; scanf("%d",&T); while(T--){ scanf("%d%d",&n,&m); for(int i = 1;i <= n;++i){ for(int j = 1;j <= m;++j){ scanf("%d",&color[i][j]); } } double numerator = 0,denominator = 0; for(int i = 1;i <= n;++i){ for(int j = 1;j <= m;++j){ denominator += i * j; numerator += calcu(i,j); } } printf("%.9lf ",numerator / denominator); } return 0; } /* 1 3 3 1 2 3 4 5 6 7 8 9 */