#include <memory.h>
#include <iomanip>
#include <cmath>
using namespace std;
#include <iostream>
int s[9][9]; // 给个各自的分数
int sum[9][9]; // (1,1)到(i,j)的矩形的分数之和
int res[15][9][9][9][9];
int calSum(int x1,int y1,int x2,int y2){//(x1,y1)到(x2,y2)的矩形的分数之和
return sum[x2][y2]- sum[x2][y1-1] - sum[x1-1][y2] + sum[x1-1][y1-1];
}
int fun(int n,int x1,int y1,int x2, int y2){
int t,a,b,c,e,MIN = 10000000;
if(res[n][x1][y1][x2][y2]!=-1)
return res[n][x1][y1][x2][y2];
if(n==1){
t = calSum(x1,y1,x2,y2);
res[n][x1][y1][x2][y2] = t*t;
return t*t;
}
for(a = x1;a<x2;a++){
c = calSum(a+1,y1,x2,y2);
e = calSum(x1,y1,a,y2);
t = min(fun(n-1,x1,y1,a,y2)+ c*c, fun(n-1,a+1,y1,x2,y2)+e*e);
if(MIN > t) MIN = t;
}
for(b = y1; b<y2; b++){
c =calSum(x1,b+1,x2,y2);
e = calSum(x1,y1,x2,b);
t = min(fun(n-1,x1,y1,x2,b)+c*c,fun(n-1,x1,b+1,x2,y2)+e*e);
if(MIN > t) MIN = t;
}
res[n][x1][y1][x2][y2] = MIN;
return MIN;
}
int main(){
memset(sum,0,sizeof(sum));
memset(res,-1,sizeof(res));
int n;
cin>>n;
for(int i = 1;i<9;i++){
for(int j=1,rowsum=0;j<9;j++){
cin>>s[i][j];
rowsum +=s[i][j];
sum[i][j] += sum[i-1][j] + rowsum;
}
}
double result = n * fun(n,1,1,8,8)-sum[8][8]*sum[8][8];
cout<<setiosflags(ios::fixed)<<setprecision(3)<<sqrt(result/(n*n))<<endl;
return 0;
}
北大 算法 2.3 递归 棋盘分割
原文作者:骑士周游问题
原文地址: https://blog.csdn.net/weixin_37629738/article/details/61919635
本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
原文地址: https://blog.csdn.net/weixin_37629738/article/details/61919635
本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。