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http://www.lydsy.com/JudgeOnline/problem.php?id=1048
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\n", ans+1e-6); return 0; }
注意到数据很小....推一下公式就爆搜吧...
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原文地址:http://www.cnblogs.com/iwtwiioi/p/4211568.html