1105 Spiral Matrix （25 分）_matrix 25-优快云博客

本文详细解析了螺旋矩阵填充算法，介绍了如何将一系列正整数按逆时针螺旋顺序填入矩阵，确保矩阵的行数大于等于列数且两者之差最小。通过实例演示了算法的实现过程，包括输入序列的排序、矩阵边界的设定及螺旋填充的步骤。

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1105 Spiral Matrix （25 分）

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:

关于螺旋矩阵的问题：可以一层一层的去解决，先将数组从大到小排序设置矩阵的边界条件，上为u. 下为d 左为l 右为r 一层一层去填写，一层填写完成后，u++ d-- l++ r-- . 最先要算出矩阵的规模，要求m行n列，m>=n m-n最小，那么m n都肯定是其因子，故可求出数字的开方的上界, 往上++，直到N%m==0 此时m-n最小。注意只有一个元素时，直接输出。

#include<iostream>
#include<string>
#include<stdio.h>
#include<string.h>
#include<vector>
#include<set>
#include<map>
#include<unordered_map>
#include<queue>
#include<algorithm>
#include<cmath>
using namespace std;
#define ll long long
#define vi vector<int>
#define pii pair<int, int>
#define x first
#define y second
const int maxn=11000;
bool cmp(int a,int b)
{
    return a>b;
}
int mar[maxn][maxn],n,ans[maxn];
int main()
{
    cin>>n;
    for(int i=0;i<n;i++) cin>>ans[i];
    if(n==1) {printf("%d",ans[0]); return 0;}
    sort(ans,ans+n,cmp);
    int m=(int)ceil(sqrt(1.0*n));
    while(n%m!=0) m++;
    int c=n/m, i=1,j=1,now=0;
    int u=1,d=m,l=1,r=c;
    while(now<n)
    {
        while(now<n&&j<r)
        {
            mar[i][j]=ans[now++];
            j++;
        }
        while(now<n&&i<d)
        {
            mar[i][j]=ans[now++];
            i++;
        }
        while(now<n&&j>l)
        {
            mar[i][j]=ans[now++];
            j--;
        }
        while(now<n&&i>u)
        {

            mar[i][j]=ans[now++];
            i--;
        }
        u++;d--;l++;r--;
        i++;j++;
        if(now==n-1)
        {

             mar[i][j]=ans[now++];
        }

    }
    for(int i=1;i<=m;i++)
    {
        for(int j=1;j<=c;j++)
        {
          if(j<c) cout<<mar[i][j]<<" ";
          else cout<<mar[i][j]<<endl;
        }
    }
    return 0;
}