PAT_A 1105. Spiral Matrix (25)

1105. Spiral Matrix (25)

• 题目信息
  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

• 简述
  其实就是讲一堆数排序后，顺时针填充二位数组，挺绕的。

• 我那超长的代码（下次总结，看以下别人是怎么做的）
  
  • 代码太长了
  • vector按序加入数据，这个忘了，以后回来再改。

#include<iostream>
#include<vector>
#include<algorithm>
#include<cmath>
int m[10000][10000];
using namespace std;
int r=0;
int c=0;
//顺时针填充数组
void ass(vector<int>&d)
{int l=0;int ii=1;int jj=1;int size=d.size()-1;int pos=1;while(d[0]!=size){if(l==0){l=1;for(int i=1;i<=c;i++){if(m[ii][i]==0){jj=i;m[ii][jj]=d[pos];pos++;d[0]++;}}}else if(l==1){l=2;for(int i=1;i<=r;i++){if(m[i][jj]==0){ii=i;m[ii][jj]=d[pos];pos++;d[0]++;}}}else if(l==2){l=3;for(int i=c;i>0;i--){if(m[ii][i]==0){jj=i;m[ii][jj]=d[pos];pos++;d[0]++;}}}else{l=0;for(int i=r;i>0;i--){if(m[i][jj]==0){ii=i;m[ii][jj]=d[pos];pos++;d[0]++;}}}}
}
bool cam(int a,int b)
{return a>b;
}
int main()
{int n;vector<int> d;vector<int> v;d.push_back(0);v.push_back(0);//需要添加数据时，直接排序cin>>n;for(int i=0;i<n;i++){int tmp;cin>>tmp;d.push_back(tmp);v.push_back(0);}int half=sqrt(n);for(int i=half;i>0;i--){if(n%i==0){half=i;break;}}c=half;r=n/c;sort(d.begin()+1,d.begin()+d.size(),cam);ass(d);for(int i=1;i<=r;i++){for(int j=1;j<=c;j++){cout<<m[i][j];if(j<c)cout<<" ";}cout<<endl;}return 0;
}