poj-1191

最新推荐文章于 2021-09-28 21:07:09 发布

fyfcauc

最新推荐文章于 2021-09-28 21:07:09 发布

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分类专栏： DP POJ 文章标签： poj

本文链接：https://blog.youkuaiyun.com/fyfcauc/article/details/38960089

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//592K	32MS	C++
#include <cstdio>
#include <cstring>
#include <cmath>

using namespace std;

long DP[16][9][9][9][9];

int chess[8][8];

long getSquareSum(int x1, int y1, int x2, int y2) {
	if (x1 > x2 || y1 < y2) {
		return 0;
	}

	long sum = 0;
	for (int i = y2; i <= y1; i++) {
		for (int j = x1; j <= x2; j++) {
			sum += chess[j][i];
		}
	}
	return sum*sum;
}

int N;

#define INF 99999999

inline long min(long a, long b) {
	return a < b ? a: b;
}

long totalSum;

void getMin() {
	memset(DP, 0, sizeof(DP));
	for (int k = 1; k <= N; k++) {
		for (int x1 = 0; x1 <= 7; x1++) {
			for (int y1 = 0; y1 <= 7; y1++) {
				for (int x2 = 0; x2 <= 7; x2++) {
					for (int y2 = 0; y2 <= 7; y2++) {
						if (k == 1) {
							DP[k][x1][y1][x2][y2] = getSquareSum(x1, y1, x2, y2);
						} else if (x2 == x1 && y1 == y2) {// impossible to cut to k parts
							// printf("%d %d %d %d %d\n", k, x1, y1, x2, y2);
							DP[k][x1][y1][x2][y2] = INF;
						// } else if (x2 - x1 + 1 >= k && y1 - y2 + 1 < k) { // only can cut x
						// 	DP[k][x1][y1][x2][y2] = INF;
						// 	for (int cutPos = x1; cutPos <= x2; cutPos++) {
						// 		long chose1 = getSquareSum(x1,y1, cutPos, y2) + DP[k-1][cutPos+1][y1][x2][y2];
						// 		long chose2 = getSquareSum(cutPos+1, y1, x2, y2) + DP[k-1][x1][y1][cutPos][y2];
						// 		long tmp = min(chose1, chose2);
						// 		DP[k][x1][y1][x2][y2] = min(tmp, DP[k][x1][y1][x2][y2]);
						// 	}
						// } else if (x2 - x1 < k && y1 - y2 >= k) { // only can cut y
						// 	DP[k][x1][y1][x2][y2] = INF;
						// 	for (int cutPos = y2; cutPos <= y1; cutPos++) {
						// 		long chose1 = getSquareSum(x1, y1, x2, cutPos) + DP[k-1][x1][cutPos-1][x2][y2];
						// 		long chose2 = getSquareSum(x1, cutPos-1, x2, y2) + DP[k-1][x1][y1][x2][cutPos];
						// 		long tmp = min(chose1, chose2);
						// 		DP[k][x1][y1][x2][y2] = min(tmp, DP[k][x1][y1][x2][y2]);	
						// 	}
						} else {
							DP[k][x1][y1][x2][y2] = INF;
							for (int cutPos = x1; cutPos <= x2; cutPos++) {
								long chose1 = getSquareSum(x1,y1, cutPos, y2) + DP[k-1][cutPos+1][y1][x2][y2];
								long chose2 = getSquareSum(cutPos+1, y1, x2, y2) + DP[k-1][x1][y1][cutPos][y2];
								long tmp = min(chose1, chose2);
								DP[k][x1][y1][x2][y2] = min(tmp, DP[k][x1][y1][x2][y2]);
							}
							for (int cutPos = y2; cutPos <= y1; cutPos++) {
								long chose1 = getSquareSum(x1, y1, x2, cutPos) + DP[k-1][x1][cutPos-1][x2][y2];
								long chose2 = getSquareSum(x1, cutPos-1, x2, y2) + DP[k-1][x1][y1][x2][cutPos];
								long tmp = min(chose1, chose2);
								DP[k][x1][y1][x2][y2] = min(tmp, DP[k][x1][y1][x2][y2]);	
							}
						}
					}
				}
			}
		}		
	}
	// printf("%ld %ld\n", DP[N][0][7][7][0], DP[1][0][7][7][0]);
	// printf("%ld\n", DP[N][0][7][7][0]);
	double ans=sqrt(DP[N][0][7][7][0]*1.0/N-((totalSum*1.0)/N*(totalSum*1.0)/N));
	printf("%.3f\n", ans);
}

int main() {
	scanf("%d", &N);
	totalSum = 0;
	for (int i = 0; i <= 7; i++) {
		for (int j = 0; j <= 7; j++) {
			scanf("%d", &chess[j][i]);
			// printf("%d\n", chess[j][i]);
			totalSum += chess[j][i];
		}
	}
	getMin();
}

http://blog.youkuaiyun.com/xymscau/article/details/6854410

转化稍微有些难度，主要是要将方程变下形，然后才能看出来：

S^2 = 1/n *（∑(Xi - X)^2）

= 1/n(n*X^2 + ∑Xi^2 - 2*X∑Xi)

= 1/n（∑Xi^2） - X^2

这里的Xi是每个切割出来的矩形的矩形和，X是整个大矩形的矩形和/(8*8) , X 在输入完数据以后就确定了，那么为了使S尽量小，就要尽量使Xi(1~N)的平方和的和尽量小了，

一个重要的理解是，题目最后要得到N个矩形，那么一共要切 N - 1刀，注意的是，一个矩形被从大矩形切出来以后，就不能再被切割了，而是作为最后的N个矩形中的一块存在，这一点很重要，否则，如果之前切出来的矩阵也可以被切，就得不到下面的状态方程了：

DP[K][x1][y1][x2][y2]:

指的是对于左上角是(x1, y1), 右下角是(x2, y2)的矩阵部分进行切割，得到K块子矩阵时， K个矩阵的矩阵和的平方的和的最小值，

那么显然，题目要求其实就是 DP[N][0][8][8][0]这个值，

递归如下：

<case 1>: 如果K== 1, 那么不需要再切割，只有一个矩阵，直接返回(x1, y1, x2, y2)所代表的矩阵和的平方即可。