UVALive 7269 Snake Carpet (构造)

题目：传送门。

题意:构造出一个矩阵,使得矩阵含有n条蛇,每条蛇的长度是1到n,并且奇数长度的蛇有奇数个拐弯,偶数长度

的蛇有偶数个拐弯.

奇数和偶数分开构造,奇数可以是:

1357

3357

5557

7777

这样一直构造下去,偶数可以这样:

2266

4466

4466

8888

8888

不断的放右边和放下面.

然后预处理每一个蛇的每一个坐标,然后考虑一下偶数的是放在奇数部分的右边还是下面.

#include <bits/stdc++.h>
using namespace std;
#define maxn 1111

struct point {
    int x, y;
};
vector <point> p[maxn];
int n;

void init () {
    for (int i = 1; i <= 1000; i++) {
        p[i].clear ();
    }
    for (int i = 1; i <= 1000; i++) {
        if (i&1) {
            int x = (i+1)/2;
            int y = x;
            for (int j = 1; j <= x; j++) {
                p[i].push_back ((point) {x, j});
            }
            for (int j = x-1; j >= 1; j--) {
                p[i].push_back ((point) {j, y});
            }
        }
        else {
            if (i == 2) {
                p[2].push_back ((point) {1, 1});
                p[2].push_back ((point) {1, 2});
            }
            else if ((i>>1)&1) {
                int y = i>>1;
                for (int j = 1; j <= y; j++) {
                    p[i].push_back ((point) {j, y});
                }
                for (int j = y; j >= 1; j--) {
                    p[i].push_back ((point) {j, y+1});
                }
            }
            else {
                int x = i>>1;
                for (int j = 1; j <= x; j++) {
                    p[i].push_back ((point) {x, j});
                }
                for (int j = x; j >= 1; j--) {
                    p[i].push_back ((point) {x+1, j});
                }
            }
        }
    }
}

int main () {
    init ();
    while (scanf ("%d", &n) == 1) {
        if (((n+1)/2)&1) {//偶数放在右边
            printf ("%d %d\n", (n+1)>>1, (n/2)+(n/2)+1);
            for (int i = 1; i <= n; i++) {
                for (int j = 0; j < p[i].size (); j++) {
                    if (i&1) {
                        printf ("%d %d%c", p[i][j].x, p[i][j].y, j == p[i].size()-1 ? ‘\n‘:‘ ‘);
                    }
                    else {
                        printf ("%d %d%c", p[i][j].x, p[i][j].y+((n+1)>>1), j == p[i].size()-1 ? ‘\n‘:‘ ‘);
                    }
                }
            }
        }
        else {//偶数放在下面
            printf ("%d %d\n", (n/2)+(n/2)+1, (n+1)>>1);
            for (int i = 1; i <= n; i++) {
                for (int j = 0; j < p[i].size (); j++) {
                    if (i&1) {
                        printf ("%d %d%c", p[i][j].x, p[i][j].y, j == p[i].size()-1 ? ‘\n‘:‘ ‘);
                    }
                    else {
                        printf ("%d %d%c", p[i][j].x+((n+1)>>1), p[i][j].y, j == p[i].size()-1 ? ‘\n‘:‘ ‘);
                    }
                }
            }
        }
    }
    return 0;
}