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
- 思路1：按规则将数组存入二维矩阵再输出
step 1: 行和列的获取，行 >= 列，因为sqrt(n) * sqrt(n) = n，不可能行和列都比sqr小，否则乘积必<n； 故行>=sqr，列<=sqr, 可从sqr向下枚举找到第一个可被n整除的数，即为列，向上即为行（求出列后，直接用n / col即可）
step 2: 先将第一个元素存入，然后每走一步提前判断是否合法（越界否？是否已有元素？），若合法走下一步，否则顺时针转90度（用四个循环来控制）
- code:
#include <bits/stdc++.h>
using namespace std;
int in[11000], ans[11000][11000];
bool T(int x, int y, int c, int r){
return ans[y][x] != 0 || x < 1 || x > c || y < 1 || y > r ? false : true;
}
int Getn(int n){
int sqr = sqrt(1.0 * n);
while(n % sqr != 0){
sqr--;
}
return sqr;
}
void Print(int r, int c){
for(int i = 1; i <= r; ++i){
for(int j = 1; j <= c; ++j){
if(j == 1) printf("%d", ans[i][j]);
else printf(" %d", ans[i][j]);
}
printf("\n");
}
}
int main(){
int n;
scanf("%d", &n);
for(int i = 0; i < n; ++i) scanf("%d", &in[i]);
int col = Getn(n);
int row = n / col;
int x = 1, y = 1, idex = 0;
sort(in, in+n, greater<int>());
ans[y][x] = in[idex++];
while(idex < n){
while(idex < n && T(x+1, y, col, row)){
ans[y][++x] = in[idex++];
}
while(idex < n && T(x, y+1, col, row)){
ans[++y][x] = in[idex++];
}
while(idex < n && T(x-1, y, col, row)){
ans[y][--x] = in[idex++];
}
while(idex < n && T(x, y-1, col, row)){
ans[--y][x] = in[idex++];
}
}
Print(row, col);
return 0;
}
- 思路2： 优先队列+增量数组+ctl控制
纯粹是想练练优先队列的写法（用的不多），排序就可以
通过增量数组控制(x, y)，随着ctl++，方向变化为：右，下，左，上…不断重复， 为了实现这个不断重复的效果，ctrl % 4；这样就实现了每次“碰壁”，旋转90度的效果。
步骤：将元素压入优先队列，初始 (x，y) = (1, 1)；
每次取队首，存入G[x][y]中，若下一步“碰壁”，更新ctl = (ctl + 1) % 4;（旋转90度）直到队空结束，输出二维数组
- T2 code
#include <bits/stdc++.h>
using namespace std;
const int maxn = 10010;
int G[maxn][maxn];
int X[5] = {0, 1, 0, -1}; //左、下、右、上
int Y[5] = {1, 0, -1, 0};
bool Judge(int x, int y, int r, int c){
return G[x][y] != 0 || x < 1 || x > r || y < 1 || y > c ? true : false;
}
int main(){
int n;
scanf("%d", &n);
priority_queue<int, vector<int>, less<int> > pq;
for(int i = 0; i < n; ++i){
int tmp;
scanf("%d", &tmp);
pq.push(tmp);
}
int col = sqrt(1.0 * n);
while(n % col != 0){ //Wrong 1:
col--;
}
int row = n / col, x = 1, y = 1, ctl = 0;
while(!pq.empty()){
G[x][y] = pq.top();
pq.pop();
if(Judge(x + X[ctl], y + Y[ctl], row, col)){
ctl = (ctl + 1) % 4;
}
x = x + X[ctl];
y = y + Y[ctl];
}
for(int i = 1; i <= row; ++i){
for(int j = 1; j <= col; ++j){
printf("%d", G[i][j]);
if(j < col) printf(" ");
}
printf("\n");
}
return 0;
}
