构造 hihocoder 1257 Snake Carpet (15北京I)-优快云博客

本文详细解析了贪吃蛇构造问题，包括构造规律、具体实例及代码实现，涉及矩阵矩形相关知识，适用于算法与数据结构学习。

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题目传送门

题意：贪吃蛇，要求长度奇数的蛇转弯次数为正奇数，长度偶数转弯次数为正偶数，且组成矩形。(北大出的题咋都和矩形相关！！！)

分析：构造找规律，想到就简单了。可以构造宽：(n + 1) / 2，长(n + 1) * n / 2 / (n + 1) / 2的矩形;

n = 5

1 2 4 4 5
3 2 4 4 5
3 3 5 5 5

n = 7

1 2 4 4 5 6 6
3 2 4 4 5 6 6
3 3 5 5 5 7 6
7 7 7 7 7 7 6

n = 8

1 2 4 4 5 6 6 8 8
3 2 4 4 5 6 6 8 8
3 3 5 5 5 7 6 8 8
7 7 7 7 7 7 6 8 8

n = 9

1 2 4 4 5 6 6 8 8
3 2 4 4 5 6 6 8 8
3 3 5 5 5 7 6 8 8
7 7 7 7 7 7 6 9 8
9 9 9 9 9 9 9 9 8

#include <bits/stdc++.h>

using namespace std;



struct Point {

	int x, y;

	Point () {}

	Point (int x, int y) : x (x), y (y) {}

};

char ans[5][100] = { "1 1\n1 1\n", "1 3\n1 1\n1 2 1 3\n", "2 3\n1 2\n1 3 2 3\n1 1 2 1 2 2\n",

                     "2 5\n1 4\n1 5 2 5\n1 1 2 1 2 2\n1 2 1 3 2 3 2 4\n",

                     "3 4\n1 4 1 5\n2 4 2 5 3 5\n2 2 2 3 3 3 3 2\n3 1 2 1 1 1 1 2 1 3\n"};



void print(vector<Point> &vec)	{

	printf ("%d %d", vec[0].x, vec[0].y);

	for (int i=1; i<vec.size (); ++i)	{

		printf (" %d %d", vec[i].x, vec[i].y);

	}

	puts ("");

	vec.clear ();

}



void DFS(int r, int c, int d, int n)	{

	if (d > n)	return ;

	vector<Point> vec;

	if (d == n)	{

		for (int i=1; i<r; ++i)	vec.push_back (Point (i, c));

		for (int i=r-1; i>=1; --i)	vec.push_back (Point (i, c + 1));

		print (vec);

		return ;

	}

	else	{

		for (int i=r-2; i>=1; --i)	vec.push_back (Point (i, c));

		for (int i=1; i<=r; ++i)	vec.push_back (Point (i, c + 1));

		print (vec);

		for (int i=1; i<=c; ++i)	vec.push_back (Point (r, i));

		vec.push_back (Point (r-1, c));

		print (vec);

	}

	DFS (r + 1, c + 2, d + 2, n);

}



int main(void)	{

	int n;

	while (scanf ("%d", &n) == 1)	{

		if (n < 5)	{

			printf ("%s", ans[n-1]);

		}

		else	{

			int x = (n + 1) / 2;

			int y = (n + 1) * n / 2 / x;

			printf ("%d %d\n", x, y);

			printf ("%s", ans[4]);

			DFS (4, 6, 6, n);

		}

	}



	return 0;

}

转载于:https://www.cnblogs.com/Running-Time/p/4978194.html