poj1191——分割棋盘求最小均方差,记忆化搜索

我的记忆化搜索啊。。。这貌似不是我写的代码555，我的找不到了

#include <iostream>

#include <cmath>

using namespace std;

#define MIN(a, b) a>b?b:a

const int size = 9;

const int nMax = 15;

const int inf = 10000000; //我有个小疑惑，不能将inf初始化为0x7fffffff和INT_MAX

int dp[nMax][size][size][size][size];

int sum[size][size]; //记录的是左上角为（1，1）的所有矩形的和

int getSum(int x1, int y1, int x2, int y2) //根据数组sum存储的值求任意矩形的和

{

return sum[x2][y2]-sum[x1-1][y2]-sum[x2][y1-1]+sum[x1-1][y1-1];

}

int cut(int k, int x1, int y1, int x2, int y2)

{

int s1, s2;

if(dp[k][x1][y1][x2][y2] != -1) //已经求过就可以直接返回了，避免重复运算

return dp[k][x1][y1][x2][y2];

if(k == 1) // k == 1说明切割了n-1次，最后剩下的一块矩形求出其和的平方直接返回

{

s1 = getSum(x1, y1, x2, y2);

return (dp[k][x1][y1][x2][y2] = s1*s1);

}

//水平方向切

int min = inf, tmp1; // min记录横向和纵向切割后，值较小的

for(int x = x1; x < x2; x++)

{

s1 = getSum(x1, y1, x, y2);

s2 = getSum(x+1, y1, x2, y2);

tmp1 = MIN(cut(k-1, x+1, y1, x2, y2) + s1*s1, cut(k-1, x1, y1, x, y2) + s2*s2);

//横向将当前的矩形切割为两块后，继续将两块进行切割，最后返回时，选取得到的值较小的

if(tmp1 < min)

min = tmp1;

}

//竖直方向切

int tmp2;

for(int y = y1; y < y2; y++)

{

s1 = getSum(x1, y1, x2, y);

s2 = getSum(x1, y+1, x2, y2);

tmp2 = MIN(cut(k-1, x1, y+1, x2, y2) + s1*s1, cut(k-1, x1, y1, x2, y) + s2*s2);//与横向切割类似

if(tmp2 < min)

min = tmp2;

}

return (dp[k][x1][y1][x2][y2] = min);

}

int main()

{

int i, j, val, total, n;

double avrg;

while(scanf("%d", &n) != EOF)

{

memset(dp, -1, sizeof(dp));

memset(sum, 0, sizeof(sum)); //必须把sum初始化为0

total = 0;

for(i = 1; i < size; i++)//注意i，j从1开始

for(j = 1; j < size; j++)

{

scanf("%d", &val);

total += val;

sum[i][j] = val + sum[i][j-1] + sum[i-1][j] - sum[i-1][j-1]; //注意i，j为1时的处理，存储sum[i][j]

}

avrg = total*1.0/n;

int res = cut(n, 1, 1, size-1, size-1);

double ans = sqrt(res*1.0/n-avrg*avrg);

printf("%.3lf\n", ans);

//system("pause");

}

return 0;

}