【BZOJ】1048: [HAOI2007]分割矩阵

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

题意：给出一个a×b（a,b<=10）的矩阵，带一个<=100的权值，现在要切割n-1次变成n个矩形（n<=10），求

$$sqrt{frac{1}{n}sum_{i=1}^{n}(sum[i]-mu)}, mu = frac{sum_{i=1}^{n} sum[i]}{n}, sum[i]表示矩阵的和$$

的最小值

#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; static 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 now=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=min(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, kk)+dfs(x1, i+1, x2, y2, k-kk));
	return now;
}

int main() {
	init();
	double ans=(double)dfs(1, 1, A, B, n-1)-(double)sqr(sum[A][B])/(double)n;
	ans=sqrt((double)1/n)*sqrt(ans);
	printf("%.2f
", ans+1e-6);
	return 0;
}

　　

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注意到数据很小....推一下公式就爆搜吧...