PAT (Advanced Level) Practise 1105 Spiral Matrix (25)

1105. Spiral Matrix (25)

时间限制

150 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

This time your job is to fill a sequence of N positive integers into a spiral matrix in non-increasing order. A spiral matrix is filled in from the first element at the upper-left corner, then move in a clockwise spiral. The matrix has m rows and ncolumns, where m and n satisfy the following: m*n must be equal to N; m>=n; and m-n is the minimum of all the possible values.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N. Then the next line contains N positive integers to be filled into the spiral matrix. All the numbers are no more than 10⁴. The numbers in a line are separated by spaces.

Output Specification:

For each test case, output the resulting matrix in m lines, each contains n numbers. There must be exactly 1 space between two adjacent numbers, and no extra space at the end of each line.

Sample Input:

1237 76 20 98 76 42 53 95 60 81 58 93

Sample Output:

98 95 9342 37 8153 20 7658 60 76

题意：将N个数从大到小按蛇形的方式放入m行n列的矩阵，并且m*n=N，m>n，m-n最小

解题思路：先从小到大排序，然后按蛇形的方向（右下左上）填入矩阵

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <algorithm>#include <queue>#include <vector>#include <set>#include <stack>#include <map>#include <climits>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;int a[10005],ans[1000][1000];int n, r, c;int dir[4][2] = {{-1,0},{0,1},{1,0},{ 0,-1 }};int main(){while (~scanf("%d", &n)){int c = sqrt(n);while (n%c != 0) c--;r = n / c;for (int i = 1; i <= n; i++) scanf("%d", &a[i]);sort(a + 1, a + 1 + n,greater<int>());memset(ans, 0, sizeof ans);int x = 1, y = 1,p=0;for (int i = 1; i <= n; i++){ans[x][y] = a[i];if (i == n) break;while (1){int xx = x + dir[p][0];int yy = y + dir[p][1];if (ans[xx][yy] || xx<1 || xx>r || yy<1 || yy>c) { p++; p %= 4; continue; }x = xx, y = yy; break;}}for (int i = 1; i <= r; i++){printf("%d", ans[i][1]);for (int j = 2; j <= c; j++)printf(" %d", ans[i][j]);printf("\n");}}return 0;}