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 1. 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

排序后利用一个dir进行上下左右的方向切换即可。

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int N;
int main() {
    scanf("%d", &N);
    vector<int> v(N);
    for(int i = 0; i < N; i++)
        scanf("%d", &v[i]);
    sort(v.begin(), v.end(), greater<int>());
    int col, row;
    for(int i = 1; i * i <= N; i++)
        if(N % i == 0) col = i;
    row = N / col;
    vector<vector<int>> ans(row);
    for(int i = 0; i < row; i++) ans[i].resize(col, -1);
    int x = 0, y = 0, dir = 0, i = 0;// 0 右边 1 下面 2 左面 3 上面
    if(y + 1 >= col || ans[x][y + 1] != -1) dir = (dir + 1) % 4; // 开始方向可能朝下
    while(true) {
        if(i == v.size()) break;
        if(dir == 0) {
            ans[x][y++] = v[i++]; 
            if(y + 1 >= col || ans[x][y + 1] != -1) dir = (dir + 1) % 4;
        } else if(dir == 1) {
            ans[x++][y] = v[i++];
            if(x + 1 >= row || ans[x + 1][y] != -1) dir = (dir + 1) % 4;
        } else if(dir == 2) {
            ans[x][y--] = v[i++];
            if(y - 1 < 0 || ans[x][y - 1] != -1) dir = (dir + 1) % 4;
        } else {
            ans[x--][y] = v[i++];
            if(x - 1 < 0 || ans[x - 1][y] != -1) dir = (dir + 1) % 4;
        }
    }
    for(int i = 0; i < row; i++) {
        printf("%d", ans[i][0]);
        for(int j = 1; j < col; j++) 
            printf(" %d", ans[i][j]);
        putchar('\n');
    }
    return 0;
}