hihoCoder 1257

In school of EECS of Peking University, there is a homework for all freshman — the contest of AI
snakes. This contest is ended today. Bacchus has got a very good result, so he decides to make a carpet
full of snakes as a souvenir, and lays it over the floor in his room.
As his room is square, a square carpet is needed. A H × W carpets is made up of H × W units
(each unit is 1 × 1). Snakes can have different length, but all snakes’ width is 1 unit. For some reason,
he hopes that N special snakes are drawn on the carpet: the length of the i-th snake should be i, which
can be seen as i connected units (Two units that share an edge are considered connected). Except the
first snake, the (2k−1)-th snake should have positive odd number of turning points; except the second
snake, the 2k-th snake should have an positive even number of turning points. i and k both start from
1. Each snake should not intersect with itself, nor with other snakes. All units of the carpet must be
covered by snakes.
But the question is whether there is a solution.
Input
Multiple test cases. There will be up to 25 cases.
For each test case: one line contains one integer N, indicating the number of snakes. (1 ≤ N ≤ 500)
Output
For each test case:
If the solution does not exist, output one line ‘0 0’, otherwise output N + 1 lines: The first line
contains two integers H and W, indicating the height and the width of the carpet. You should guarantee
that H × W = 1 + 2 + . . . + N. For the next N lines, the i-th line contain 2i integers, indicating the
coordinates of the i-th snake in order. The coordinate of top-left corner unit is (1, 1) and the coordinate
of bottom-right corner unit is (H, W).
Hint: This problem is special judged, and the solutions for the
sample input are on the right:
Sample Input
3
4
5
Sample Output
2 3
1 2
1 3 2 3
1 1 2 1 2 2
2 5
1 4
1 5 2 5
1 1 2 1 2 2
1 2 1 3 2 3 2 4
3 5
3 4
1 4 1 5
2 4 2 5 3 5
2 2 2 3 3 3 3 2

3 1 2 1 1 1 1 2 1 3

简单构造

#include 
#include
#include
#include
#include
using namespace std;
const int maxn=505;
struct node
{
    int x,y;
    void init(int x1,int y1)
    {
        x=x1,y=y1;
    }
};
node g[maxn][maxn];
void init()
{
    memset(g,0,sizeof(g));
    g[1][1].init(3,4),g[2][1].init(1,4),g[2][2].init(1,5);
    g[3][1].init(2,4),g[3][2].init(2,5),g[3][3].init(3,5);
    g[4][1].init(2,2),g[4][2].init(2,3),g[4][3].init(3,3),g[4][4].init(3,2);
    g[5][1].init(3,1),g[5][2].init(2,1),g[5][3].init(1,1),g[5][4].init(1,2),g[5][5].init(1,3);
}
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        init();
        if(n==1)
        {
            printf("1 1\n1 1\n");
            continue;
        }
        if(n==2)
        {
            printf("1 3\n1 1\n1 2 1 3\n");
            continue;
        }
        if(n==3)
        {
            printf("2 3\n1 2\n1 3 2 3\n1 1 2 1 2 2\n");
            continue;
        }
        if(n==4)
        {
            printf("2 5\n1 4\n1 5 2 5\n1 1 2 1 2 2\n1 2 1 3 2 3 2 4\n");
            continue;
        }
        if(n==5)
        {
            printf("3 5\n3 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");
            continue;
        }
        for(int x=6; x<=n; x++)
        {
            int h=(x+1)/2,w=(x+1)*x/2/h;
            int h1=x/2,w1=(x-1)*x/2/h1;
            int cnt=1;
            if(x%2==0)
            {
                for(int i=1; i<=h; i++)
                {
                    g[x][cnt].init(i,w1+1);
                    cnt++;
                }
                for(int i=h;i>=1;i--)
                    g[x][cnt].init(i,w),cnt++;
            }
            else
            {
                for(int i=1; i<=x-1; i++)
                {
                    g[x][cnt].init(h,i);
                    cnt++;
                }
                g[x][cnt].init(h-1,x-1);
                cnt=1;
                for(int i=h;i>=1;i--)
                    g[x-1][cnt].init(i,w),cnt++;
                for(int i=1;i<=h-2;i++)
                    g[x-1][cnt].init(i,w-1),cnt++;
            }
        }
        int h=(n+1)/2,w=(n+1)*n/2/h;
        printf("%d %d\n",h,w);
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=i;j++)
            {
                if(j==1) printf("%d %d",g[i][j].x,g[i][j].y);
                else printf(" %d %d",g[i][j].x,g[i][j].y);
            }
            printf("\n");
        }
    }
    return 0;
}