HDU2074 叠筐

叠筐

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16016    Accepted Submission(s): 4184

Problem Description

需要的时候，就把一个个大小差一圈的筐叠上去，使得从上往下看时，边筐花色交错。这个工作现在要让计算机来完成，得看你的了。

 

Input

输入是一个个的三元组，分别是，外筐尺寸n（n为满足0<n<80的奇整数），中心花色字符，外筐花色字符，后二者都为ASCII可见字符；

 

Output

输出叠在一起的筐图案，中心花色与外筐花色字符从内层起交错相叠，多筐相叠时，最外筐的角总是被打磨掉。叠筐与叠筐之间应有一行间隔。

 

Sample Input

11 B A
5 @ W

 

Sample Output

 AAAAAAAAA 
ABBBBBBBBBA
ABAAAAAAABA
ABABBBBBABA
ABABAAABABA
ABABABABABA
ABABAAABABA
ABABBBBBABA
ABAAAAAAABA
ABBBBBBBBBA
 AAAAAAAAA 

 @@@ 
@WWW@
@W@W@
@WWW@
 @@@ 

 

题目很直白，刚开始是用的蛇形填空，结果超时了。

给出超时代码：

#include<iostream>
using namespace std;
int main()
{
    int n;
    char a, b;
    char map[85][85];
    int count = 0;
    while (~scanf("%d%s%s", &n, &a, &b))
    {
        if (count != 0)cout << endl;
        count++;
        if (n == 1)
        {
            cout << a << endl;
            continue;
        }
        //memset(map, 0, sizeof(map));
        int x = n / 2, y = n / 2;
        int co = n / 2;
        map[x][y] = a;
        //cout << x << y << endl;
        int t = x;
        int count;
        int p = 1;
        while (t--)
        {
            count = 0;
            while (count < 8 * p)
            {
                if (p % 2 == 1)//b
                {
                    while (x + 1 <= co + p&&map[x + 1][y] != b)
                        map[++x][y] = b, count++;
                    while (y - 1 >= co - p&&map[x][y - 1] != b)
                        map[x][--y] = b, count++;
                    while (x - 1 >= co - p&&map[x - 1][y] != b)
                        map[--x][y] = b, count++;
                    while (y + 1 <= co + p&&map[x][y + 1] != b)
                        map[x][++y] = b, count++;
                }
                else//a
                {
                    while (x + 1 <= co + p&&map[x + 1][y] != a)
                        map[++x][y] = a, count++;
                    while (y - 1 >= co - p&&map[x][y - 1] != a)
                        map[x][--y] = a, count++;
                    while (x - 1 >= co - p&&map[x - 1][y] != a)
                        map[--x][y] = a, count++;
                    while (y + 1 <= co + p&&map[x][y + 1] != a)
                        map[x][++y] = a, count++;
                }
            }
            p++;
        }
        map[0][0] = ' ';
        map[0][n - 1] = ' ';
        map[n - 1][0] = ' ';
        map[n - 1][n - 1] = ' ';
        //cout << map[n - 1][n - 1] << endl;
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < n; j++)
            {
                printf("%c", map[i][j]);
            }
            printf("\n");
        }
    }
    return 0;
}

给出AC 代码：

#include<iostream>
using namespace std;
int main()
{
	int n, s1, s2, y1, y2;
	char a, b;
	char map[85][85];
	int count = 0;
	while (~scanf("%d%s%s", &n, &a, &b))
	{
		if (count != 0)cout << endl;
		count++;
		if (n == 1)
		{
			cout << a << endl;
			continue;
		}
		//memset(map, 0, sizeof(map));
		int x = n / 2, y = n / 2;
		int co = n / 2;
		map[x][y] = a;
		//cout << x << y << endl;
		int t = x;
		int count;
		//int p = 1;
		for (int p = 1; p <= co;p++)
		{
			s1 = co - p;
			s2 = co + p;
			y1 = co - p;
			y2 = co + p;
			//cout << s1 << " " << s2 << " " << y1 << " " << y2 << " " << endl;
			if (p % 2 == 1)//b
			{
				for (int i = s1; i <= s2; i++)
				{
					map[y1][i] = b;
					map[y2][i] = b;
				}
				for (int i = y1+1; i <= y2; i++)
				{
					map[i][s1] = b;
					map[i][s2] = b;
				}
				
			}
			else //a
			{
				for(int i = s1; i <= s2; i++)
				{
					map[y1][i] = a;
					map[y2][i] = a;
				}
				for (int i = y1 + 1; i <= y2; i++)
				{
					map[i][s1] = a;
					map[i][s2] = a;
				}
			}
			//cout << p << endl;
		}
		map[0][0] = ' ';
		map[0][n - 1] = ' ';
		map[n - 1][0] = ' ';
		map[n - 1][n - 1] = ' ';
		//cout << map[n - 1][n - 1] << endl;
		for (int i = 0; i < n; i++)
		{
			for (int j = 0; j < n; j++)
			{
				printf("%c", map[i][j]);
			}
			printf("\n");
		}
	}
	return 0;
}