【编程题解】Workaholic Xiaoming

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本文介绍了一个基于无限网格的填充算法，通过执行一系列操作来填充单元格，并计算填充后的总单元格数量。算法使用Java实现，详细展示了如何处理行和列的填充操作，以及如何避免重复填充同一单元格。

Description

XiaoMing is a workaholic and has a lot of things to do, so he has an infinite grid pattern. In the initial state, all the grids are blank.

Xiaoming then performs n operations, each of which selects a row or a column, and selects two end point grids on that row or column, populating all the grids in the interval with the two end points with what to do (including the two end points).

How many cells will be filled after n operations? The same cell will be filled only once.

Input

The first line contains a positive integer n.

The next n rows, each containing four integers x1,y1,x2, and y2, represent the lattice coordinates of an operation.

If x1 is equal to x2, it fills a column, and if y1 is equal to y2, it fills a row.

Ensure that each operation is stained on one row or column.

Data range:1≤n≤10000,−10^9≤x1,y1,x2,y2≤10^9

Output

Contains a positive integer representing the number of cells filled.

Sample Input 1

Sample Output 1

Personal Answer (using language：JAVA) Not necessarily right

import java.math.*;
import java.util.*;
import java.util.stream.Collectors;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int[][] input = new int[n][4];
        for (int i = 0; i < n; i++) {
            for(int j =0; j<4;j++){
                input[i][j]=scanner.nextInt();
            }
        }

        class Grid{
            private int x;
            private int y;

            public Grid(int x, int y) {
                this.x = x;
                this.y = y;
            }

            @Override
            public boolean equals(Object obj)
            {
                Grid tempGrid= (Grid) obj;
                if (x==tempGrid.x&&y==tempGrid.y) return true;
                else return false;
            }
            @Override
            public int hashCode()
            {
                return new Integer(x).hashCode()+new Integer(y).hashCode();
            }
        }

        HashSet<Grid> hs =new HashSet<Grid>();

        for(int i=0;i<n;i++){
            if(input[i][0]==input[i][2]) {
                if (input[i][1] < input[i][3]) {
                    for (int j = input[i][1]; j <= input[i][3]; j++) {
                        hs.add(new Grid(input[i][0], j));
                    }
                }
                else if(input[i][1] > input[i][3]){
                    for (int j = input[i][3]; j <= input[i][1]; j++) {
                        hs.add(new Grid(input[i][0], j));
                    }
                }
                else{
                    hs.add(new Grid(input[i][0], input[i][1]));
                }
            }else{
                if (input[i][0] <input[i][2]) {
                    for (int j = input[i][0]; j <= input[i][2]; j++) {
                        hs.add(new Grid(j, input[i][1]));
                    }
                }
                else if (input[i][0] >input[i][2]){
                    for (int j = input[i][2]; j <= input[i][0]; j++) {
                        hs.add(new Grid(j, input[i][1]));
                    }
                }
                else{
                    hs.add(new Grid(input[i][0], input[i][1]));
                }
            }
        }

        System.out.println(hs.size());
    }
}

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