POJ1191 棋盘分割

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

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[][];    //每个格子的分数
int sum[][];    //(1,1)到(i,j)的矩形的分数之和
int res[][][][][];    //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-]-sum[x1-][y2]+sum[x1-][y1-];
}   

int fun(int n, int x1, int y1, int x2, int y2)
{
    int t, a, b, c, e, MIN = ;
    if (res[n][x1][y1][x2][y2] != -)
        return res[n][x1][y1][x2][y2];
    if (n == )
    {
        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 + , y1, x2, y2);
        e = calSum(x1, y1, a, y2);
        t = min(fun(n - , x1, y1, a, y2) + c * c, fun(n - , a + , y1, x2, y2) + e * e);
        if (MIN > t) MIN = t;
    }
    for (b = y1; b < y2; b++)
    {
        c = calSum(x1, b + , x2, y2);
        e = calSum(x1, y1, x2, b);
        t = min(fun(n - , x1, y1, x2, b) + c * c, fun(n - , x1, b + , x2, y2) + e * e);
        if (MIN > t) MIN = t;
    }
    res[n][x1][y1][x2][y2] = MIN;
    return MIN;
}
int main() {
    memset(sum, , sizeof(sum));
    memset(res, -, sizeof(res));   //初始化记录表
    int n;
    cin>>n;
    for (int i=; i<; i++)
        for (int j=, rowsum=; j<; j++)
        {
            cin>>s[i][j];
            rowsum +=s[i][j];
            sum[i][j] += sum[i-][j] + rowsum;
        }
    double result = n*fun(n,,,,)-sum[][]*sum[][];
    cout<<setiosflags(ios::fixed)<<setprecision()<<sqrt(result/(n*n))<<endl;
    return ;
}