POJ1191 棋盘分割

Time Limit: 1000MS              Memory Limit: 10000K
Total Submissions: 16587        Accepted: 5951

题目链接：

http://poj.org/problem?id=1191

解题思路：

关键是方程式的化简，思路清晰，整个题的难度不大。

AC代码（课程PPT）

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath> 
#include<iomanip>

using namespace std;

int s[9][9];    //每个格子的分数 
int sum[9][9];    //(1,1)到(i,j)的矩形的分数之和 
int res[15][9][9][9][9];    //fun的记录表 

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));   //初始化记录表  
    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; 
}