首页 > 专题系列 > Java解POJ > POJ 1222 EXTENDED LIGHTS OUT [解题报告] Java
2013
11-09

POJ 1222 EXTENDED LIGHTS OUT [解题报告] Java

EXTENDED LIGHTS OUT

问题描述 :

In an extended version of the game Lights Out, is a puzzle with 5 rows of 6 buttons each (the actual puzzle has 5 rows of 5 buttons each). Each button has a light. When a button is pressed, that button and each of its (up to four) neighbors above, below, right and left, has the state of its light reversed. (If on, the light is turned off; if off, the light is turned on.) Buttons in the corners change the state of 3 buttons; buttons on an edge change the state of 4 buttons and other buttons change the state of 5. For example, if the buttons marked X on the left below were to be pressed,the display would change to the image on the right.



The aim of the game is, starting from any initial set of lights on in the display, to press buttons to get the display to a state where all lights are off. When adjacent buttons are pressed, the action of one button can undo the effect of another. For instance, in the display below, pressing buttons marked X in the left display results in the right display.Note that the buttons in row 2 column 3 and row 2 column 5 both change the state of the button in row 2 column 4,so that, in the end, its state is unchanged.



Note:

1. It does not matter what order the buttons are pressed.

2. If a button is pressed a second time, it exactly cancels the effect of the first press, so no button ever need be pressed more than once.

3. As illustrated in the second diagram, all the lights in the first row may be turned off, by pressing the corresponding buttons in the second row. By repeating this process in each row, all the lights in the first

four rows may be turned out. Similarly, by pressing buttons in columns 2, 3 ?, all lights in the first 5 columns may be turned off.

Write a program to solve the puzzle.

输入:

The first line of the input is a positive integer n which is the number of puzzles that follow. Each puzzle will be five lines, each of which has six 0 or 1 separated by one or more spaces. A 0 indicates that the light is off, while a 1 indicates that the light is on initially.

输出:

For each puzzle, the output consists of a line with the string: “PUZZLE #m”, where m is the index of the puzzle in the input file. Following that line, is a puzzle-like display (in the same format as the input) . In this case, 1′s indicate buttons that must be pressed to solve the puzzle, while 0 indicate buttons, which are not pressed. There should be exactly one space between each 0 or 1 in the output puzzle-like display.

样例输入:

2
0 1 1 0 1 0
1 0 0 1 1 1
0 0 1 0 0 1
1 0 0 1 0 1
0 1 1 1 0 0
0 0 1 0 1 0
1 0 1 0 1 1
0 0 1 0 1 1
1 0 1 1 0 0
0 1 0 1 0 0

样例输出:

PUZZLE #1
1 0 1 0 0 1
1 1 0 1 0 1
0 0 1 0 1 1
1 0 0 1 0 0
0 1 0 0 0 0
PUZZLE #2
1 0 0 1 1 1
1 1 0 0 0 0
0 0 0 1 0 0
1 1 0 1 0 1
1 0 1 1 0 1

解题代码:

import java.util.Scanner; 

public class Main { 

    int times; 
    int[][] puzzle; 
    static final int length = 6; 
    static final int width = 5; 
    int[][] arr; 
    int temp; 
    int req; 

    public Main() { 
        Scanner scan = new Scanner(System.in); 
        times = scan.nextInt(); 
        for (int t = 0; t < times; t++) { 
            puzzle = new int[width][length]; 
            arr = new int[width][length]; 
            for (int i = 0; i < width; i++) { 
                for (int j = 0; j < length; j++) { 
                    puzzle[i][j] = scan.nextInt(); 
                } 
            } 
            force(); 
            System.out.println("PUZZLE #" + (t + 1)); 
            for (int i = 0; i < width; i++) { 
                for (int j = 0; j < length; j++) { 
                    System.out.print(arr[i][j] + " "); 
                } 
                System.out.println(); 
            } 
        } 
    } 

    //穷举第一行数据,推导下面的数据... 
    //判断最后一行是否符合要求 
    //6列需要最多计算2^6=64次 
    public void force() { 
        boolean result = false; 
        do { 
            result = search(); 
            if (result) { 
                return; 
            } 
        } while (plus(arr[0], 0)); 
    } 

    public boolean search() { 
        for (int i = 1; i < width; i++) { 
            for (int j = 0; j < length; j++) { 
                req = puzzle[i - 1][j]; 
                temp = 0; 
                if (j - 1 >= 0) { 
                    temp += arr[i - 1][j - 1]; 
                } 
                if (j + 1 < length) { 
                    temp += arr[i - 1][j + 1]; 
                } 
                if (i - 2 >= 0) { 
                    temp += arr[i - 2][j]; 
                } 
                temp += arr[i - 1][j]; 
                if (req != (temp % 2)) { 
                    arr[i][j] = 1; 
                } else { 
                    arr[i][j] = 0; 
                } 
            } 
        } 
        for (int i = 0; i < length; i++) { 
            req = puzzle[width - 1][i]; 
            temp = 0; 
            if (i - 1 >= 0) { 
                temp += arr[width - 1][i - 1]; 
            } 
            temp += arr[width - 2][i]; 
            temp += arr[width - 1][i]; 
            if (i + 1 < length) { 
                temp += arr[width - 1][i + 1]; 
            } 
            if (req != (temp % 2)) { 
                return false; 
            } 
        } 
        return true; 
    } 

    public boolean plus(int[] a, int b) { 
        if (b == length) { 
            return false; 
        } else if (a[b] == 0) { 
            a[b]++; 
            return true; 
        } else { 
            a[b] = 0; 
            return plus(a, ++b); 
        } 
    } 

    public static void main(String[] args) { 
        new Main(); 
    } 

}

  1. #include <stdio.h>
    int main()
    {
    int n,p,t[100]={1};
    for(int i=1;i<100;i++)
    t =i;
    while(scanf("%d",&n)&&n!=0){
    if(n==1)
    printf("Printing order for 1 pages:nSheet 1, front: Blank, 1n");
    else {
    if(n%4) p=n/4+1;
    else p=n/4;
    int q=4*p;
    printf("Printing order for %d pages:n",n);
    for(int i=0;i<p;i++){
    printf("Sheet %d, front: ",i+1);
    if(q>n) {printf("Blank, %dn",t[2*i+1]);}
    else {printf("%d, %dn",q,t[2*i+1]);}
    q–;//打印表前
    printf("Sheet %d, back : ",i+1);
    if(q>n) {printf("%d, Blankn",t[2*i+2]);}
    else {printf("%d, %dn",t[2*i+2],q);}
    q–;//打印表后
    }
    }
    }
    return 0;
    }

  2. 如果两个序列的最后字符不匹配(即X [M-1]!= Y [N-1])
    L(X [0 .. M-1],Y [0 .. N-1])= MAX(L(X [0 .. M-2],Y [0 .. N-1]),L(X [0 .. M-1],Y [0 .. N-1])
    这里写错了吧。

  3. Excellent Web-site! I required to ask if I might webpages and use a component of the net web website and use a number of factors for just about any faculty process. Please notify me through email regardless of whether that would be excellent. Many thanks