动态规划 ：POJ 1191 棋盘分割

题目的要求是均方差最小，按照均方差的公式，可以转化为最后分割的每个矩形的平方和最小。设f(k, x1, y1, x2, y2)左上角为x1，y1，右下角为x2，y2的矩形在切割k次之后每一块的平方和。s[x1][y1][x2][y2]左上角为x1，y1，右下角为x2，y2的矩形的平方和。
由此可以写出状态转移方程：
f(k, x1, y1, x2, y2) = min{ min(f(k-1, x1, y1, a, y2)+s[a+1][y1][x2][y2], f(k-1, a, y1, x2, y2)+s[x1][y1][a][y2]) (x1<=a<x2) (横切), min(f(k-1, x1, y1, x2, b)+s[x1][b+1][x2][y2], f(k-1, x1, b+1, x2, y2)+s[x1][y1][x2][b]) (y1<=b<y2) (纵切)}
初始状态为：f(0, 0, 0, 7, 7) = 整块棋盘的平方和。（参考《算法艺术与信息学竞赛》P116）

#include <stdio.h>
#include <math.h>
#include <algorithm>

using namespace std;

int matrix[8][8];
int s[8][8][8][8];
int dp[15][8][8][8][8];
int cnt;
double sum;

double mmin(double a, double b)
{
return a < b ? a : b;
}

int main()
{
int x1, x2, y1, y2, k, a;
double average, sum = 0;
scanf("%d", &cnt);
for (x1=0; x1<8; x1++)
{
for (y1=0; y1<8; y1++)
{
scanf("%d", &matrix[x1][y1]);
sum += matrix[x1][y1];
}
}

average = sum/cnt;

for (x1=0; x1<8; x1++)
{
for (y1=0; y1<8; y1++)
{
for (x2=x1; x2<8; x2++)
{
sum = 0;
for (y2=y1; y2<8; y2++)
{
sum += matrix[x2][y2];
if (x2 == x1)
{
s[x1][y1][x2][y2] = sum;
}
else
{
s[x1][y1][x2][y2] = s[x1][y1][x2-1][y2] + sum;
}

dp[0][x1][y1][x2][y2] = s[x1][y1][x2][y2] * s[x1][y1][x2][y2];
}
}
}
}

int temp;

for (k=1; k<=cnt-1; k++)
{
for (x1=0; x1<8; x1++)
{
for (y1=0; y1<8; y1++)
{
for (x2=x1; x2<8; x2++)
{
for (y2=y1; y2<8; y2++)
{
dp[k][x1][y1][x2][y2] = 2000000000;
for (a=x1; a<x2; a++)
{
temp = mmin(dp[k-1][x1][y1][a][y2] + s[a+1][y1][x2][y2] * s[a+1][y1][x2][y2],
dp[k-1][a+1][y1][x2][y2] + s[x1][y1][a][y2] * s[x1][y1][a][y2]);
dp[k][x1][y1][x2][y2] = mmin(dp[k][x1][y1][x2][y2], temp);
}

for (a=y1; a<y2; a++)
{
temp = mmin(dp[k-1][x1][y1][x2][a] + s[x1][a+1][x2][y2] * s[x1][a+1][x2][y2],
dp[k-1][x1][a+1][x2][y2] + s[x1][y1][x2][a] * s[x1][y1][x2][a]);
dp[k][x1][y1][x2][y2] = mmin(dp[k][x1][y1][x2][y2], temp);
}
}
}
}
}
}

double ret = sqrt((double)dp[cnt-1][0][0][7][7] / (double)cnt - average * average);
printf("%.3f/n", ret);
return 0;
}