"Bzoj" 1048: [HAOI2007] Split matrix

http://www.lydsy.com/JudgeOnline/problem.php?id=1048

Give a AXB (a,b<=10) matrix, with a <=100 weight, now to cut n-1 times into n rectangles (n<=10), ask

$$\sqrt{\frac{1}{n}\sum_{i=1}^{n} (SUM[I]-\MU)}, \mu = \frac{\sum_{i=1}^{n} sum[i]}{n}, Sum[i] represents the and $$ of the matrix

The minimum value

#include <cstdio> #include <cstring> #include <cmath> #include <string> #include <iostream > #include <algorithm> #include <queue> #include <set> #include <map> #include <sstream >using namespace Std;typedef long long ll; #define PB Push_back#define Rep (i, n) for (int i=0; i< (n); ++i) #define For1 (i,a,n) for (int i= (a), i<= (n), ++i) #define FOR2 (i,a,n) for (int i= (a);i< (n), ++i) #define FOR3 (i,a,n) for (int i= (a); I >= (n); i) #define FOR4 (i,a,n) for (int i= (a);i> (n); i) #define CC (i,a) memset (i,a,sizeof (i)) #define READ (a) a=  Getint () #define PRINT (a) printf ("%d", a) #define DBG (x) cout << (#x) << "=" << (x) << Endl#define Error (x) (! ( x) puts ("error"): 0) #define RDM (x, i) for (int i=ihead[x]; i; i=e[i].next) inline int getint () {static int r, K; r=0,k=1; St Atic Char C; C=getchar (); for (; c< ' 0 ' | | C> ' 9 '; C=getchar ()) if (c== '-') k=-1; for (; c>= ' 0 ' &&c<= ' 9 '; C=getchar ()) r=r*10+c-' 0 '; return k*r; }const int n=11, Oo=~0u>>2;int w[n][n], f[n][n][n][n][n], A, B, N, Sum[n][n];void init () {CC (F,-1); Read (A); Read (B); read (n); For1 (I, 1, a) For1 (J, 1, b) read (w[i][j]); For1 (I, 1, a) For1 (J, 1, b) sum[i][j]=sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1]+ W[I][J];} int getsum (int x1, int y1, int x2, int y2) {return sum[x2][y2]-sum[x2][y1-1]-sum[x1-1][y2]+sum[x1-1][y1-1];} int sqr (int x) {return x*x;} int dfs (int x1, int y1, int x2, int y2, int k) {int &now=f[x1][y1][x2][y2][k];if (NOW!=-1) return now;if (k==0) return n OW=SQR (Getsum (x1, y1, x2, y2)); if (x1==x2 && y1==y2) return Now=oo;now=oo;--k;for2 (i, X1, x2) for1 (KK, 0, K) now=mi N (now, DFS (x1, y1, I, Y2, kk) +dfs (i+1, y1, x2, y2, k-kk)); For2 (i, Y1, y2) for1 (KK, 0, K) now=min (now, DFS (x1, y1, x2, I, K k) +dfs (x1, i+1, x2, y2, k-kk)); return now;} int main () {init ();d ouble ans= (Double) DFS (1, 1, A, B, n-1)-(double) sqr (Sum[a][b])/(double) n;ans=sqrt ((double) 1/n) * sqrt (ans);p rintf ("%.2f\n", ans+1e-6); return 0;}

　　

Notice that the data is small .... Push the formula and pop the search ...

"Bzoj" 1048: [HAOI2007] Split matrix