设\(A_{sum} = \sum_{i=1}^{n+m-1}A_i\)

然后 : \((n+m-1)\sum_{i=1}^{n+m-1}(A_i-A_{avg})^2 \\ = (n+m-1)\sum_{i=1}^{n+m-1}(A_i^2-2A_iA_{avg}+A_{avg}^2) \\ = A_{sum}^2 + \sum_{i=1}^{n+m-1}A_{i}^2-2A_{sum}A_{i}\)

最后退出来 ans = (n+m-1)\(\sum_{i}^(n+m-1)A_i^2 - (A_i)^2\)

然后,把方差式子变换一下，然后把和作为状态暴力dp即可

代码:

#include <iostream> #include <stdio.h> #include <sting.h> #include <algorithm> #define maxn 35 #define inf 0x3f3f3f3f using namespace std ; int s[maxn][maxn] ; int dp[maxn][maxn][60*30*2] ; int T , cas = 0 , n , m ; int main () { 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",&s[i][j]) ; } } memset(dp,0x3f,sizeof(dp)) ; f[1][0][0] = f[0][1][0] = 0 ; for(int i = 1 ; i <= n ; i ++) { for(int j = 1 ; j <= m ; j ++) { int x = (n+m-1)*s[i][j] * s[i][j] , y = (i+j-1)*30 ; for(int k = 0 ; k <= y ; k ++) { f[i][j][k+s[i][j]] = min (f[i][j][k+s[i][j]],min(f[i-1][j][k],f[i][j-1][k])+x) ; } } } for(int i = 0 ; i <= 1800 ; i ++) { ans = min(ans,f[n][m][i]-i*i) ; printf("Case #%d: %d\n",++cas,ans); } } return 0 ; }

!我可不想再考一遍辣!