【CCF CSP】 201412-2 Z-scan (100 points)

Question number:	201412-2
Question name:	zigzag scan
time limit:	2.0s
Memory limit:	256.0MB
Problem Description:	Problem Description 　　In the image coding algorithm, a given square matrix needs to be zigzag scan (Zigzag Scan). Given an n×n matrix, the process of zigzag scanning is shown in the following figure: 　　For the following 4×4 matrix, 　　1 5 3 9 　　3 7 5 6 　　9 4 6 4 　　7 3 1 3 　　zigzag scan it Then get a sequence of length 16: 　　1 5 3 9 7 3 9 5 4 7 3 6 6 4 1 3 　　Please implement a zigzag scanning program, given an n×n matrix, output zigzag scanning on this matrix the result of. input format 　　The first row of the input contains an integer n that represents the size of the matrix. 　　The second to n+1th lines of the input each contain n positive integers, separated by spaces, representing the given matrix. output format 　　Output a line containing n×n integers, separated by spaces, representing the result of zigzag scanning of the input matrix. sample input 4 1 5 3 9 3 7 5 6 9 4 6 4 7 3 1 3 Sample output 1 5 3 9 7 3 9 5 4 7 3 6 6 4 1 3 Evaluate use case size and conventions 　　1≤n≤500, the matrix elements are positive integers not exceeding 1000.

Parse

Extract the scanning direction, and determine the selected number according to the scanning direction.
According to the title, there are four directions for scanning: when encountering the boundary, it is necessary to judge the direction of the direction.

right, down, down left, up right

code

C++

#include <iostream>
#include <vector> 
#define MAX_LENGTH 501
using namespace std;

enum Direction{Bottom,Bottom_left,Right,Top_right}flag;
int grad[MAX_LENGTH][MAX_LENGTH]; 

int main(int argc, char** argv) {
    int n;
    vector<int> vec_Z;
    cin>>n;
    //输入 
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            cin>>grad[i][j];
        }
    }

    //初始化第一个位置 
    int i=1,j=1;
    flag=Right;
    vec_Z.push_back(grad[1][1]);
    //开始Z扫描 
    for(int k=1;k<n*n;k++)
    {
        switch(flag)
        {
            case Bottom:
            i++;
            vec_Z.push_back(grad[i][j]);
            //如果在左边界，则左上；右边界，则右下 
            if(j==1)
                flag=Top_right;
            else if(j==n)
                flag=Bottom_left;
            break;

            case Bottom_left:
            i++;j--;
            vec_Z.push_back(grad[i][j]);
            //到达左边界时，换方向 下 
            if(j==1&&i!=n)
                flag=Bottom;
            else if(i==n)
                flag=Right;
            break;

            case Right:
            j++;
            vec_Z.push_back(grad[i][j]);
            if(i==1)//是否在上界 
                flag=Bottom_left;       
            else if(i==n) 
                flag=Top_right; 
            break;

            case Top_right:
            i--;j++;
            vec_Z.push_back(grad[i][j]);
            //到达上界 方向为右边 
            if(i==1&&j!=n)
                flag=Right;
            else if(j==n)//到达右界 ，方向向下
                flag=Bottom; 
            break;
        } 
    }

    //输出序列 
    for(vector<int>::iterator it=vec_Z.begin();it!=vec_Z.end();it++)
        cout<<*it<<' '; 

    return 0;
}