1105 Spiral Matrix (25 分)
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:
98 95 93 42 37 81 53 20 76 58 60 76
蛇形排序 記錄下行起點 行終點 列起點 列終點
程式碼
#include<bits/stdc++.h>
using namespace std;
int digit[10005];
int maze[10001][10001];
bool cmp(int a,int b){
return a > b;
}
int main()
{
int k;
scanf("%d",&k);
int s = (int)sqrt(k);
int m, n;
for(int i = s; i >= 1; i --)
{
if(k % i == 0)
{
m = k / i;
n = i;
break;
}
}
//說明此刻 m 是它的行數 n 是它的列數
for(int i = 1; i <= k; i++)
scanf("%d", &digit[i]);
sort(digit + 1,digit + 1 + k,cmp);
int now = 0;
int bm = 1, bn = 1, em = m, en = n;
while(now < k)
{
for(int i = bn; i <= en; i++)
maze[bm][i] = digit[++now];
bm += 1;
if(now >= k) break;
for(int i = bm; i <= em; i++)
maze[i][en] = digit[++now];
en -= 1;
if(now >= k) break;
for(int i = en; i >= bn; i--)
maze[em][i] = digit[++now];
em -= 1;
if(now >= k) break;
for(int i = em; i >= bm; i--)
maze[i][bn] = digit[++now];
bn += 1;
if(now >= k) break;
}
for(int i = 1; i <= m; i ++){
for(int j = 1; j <= n; j ++)
printf("%d%c",maze[i][j]," \n"[j == n]);
}
return 0;
}