1105. Spiral Matrix (25) 排序、模拟

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 n columns, 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:
12
37 76 20 98 76 42 53 95 60 81 58 93
Sample Output:
98 95 93
42 37 81
53 20 76
58 60 76
更新一个更好的方法。
首先要求出矩阵的行和列，行要大于等于列。
对数组进行排序。
建立一个矩阵数组，用于储存矩阵结果。
设四个变量minR，maxR，minC，maxC，用于记录当前需要填充的行列的边界。
填充方法按题目所述，设置当前行列，nowR，nowC，从左至右，从上到下，从右到左，从下到上进行填充。并更新行列边界值。

//// Created by aleafall on 16-12-3.//#include <iostream>#include <algorithm>#include <cmath>using namespace std;int main(){    int N, m, n;    cin >> N;    int a[N + 1];    for (int i = 0; i < N; ++i) {        scanf("%d", &a[i]);    }    sort(a, a + N);    m = (int) sqrt(1.0 * N);    while (N%m) {        ++m;    }    n = N / m;    if (n > m) {        swap(m, n);    }    int num[m + 1][n + 1];    int index = N - 1;    int minR = 0, maxR = m - 1, minC = 0, maxC = n - 1;    while (index >= 0) {        for (int nowC = minC,nowR=minR; nowC <=maxC ; ++nowC) {            num[nowR][nowC] = a[index--];        }        ++minR;        for (int nowR = minR,nowC=maxC; nowR <=maxR ; ++nowR) {            num[nowR][nowC] = a[index--];        }        --maxC;        if (index < 0) {    //数据可能已经完成了填充            break;        }        for (int nowC = maxC,nowR=maxR; nowC >=minC ; --nowC) {            num[nowR][nowC] = a[index--];        }        --maxR;        for (int nowR = maxR,nowC=minC; nowR >=minR ; --nowR) {            num[nowR][nowC] = a[index--];        }        ++minC;    }    for (int i = 0; i < m; ++i) {        for (int j = 0; j < n; ++j) {            if (j < n - 1) {                printf("%d ", num[i][j]);            } else {                printf("%d\n", num[i][j]);            }        }    }    return 0;}