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java - 使用广度优先填充算法填充 "power grid";我被困住了

转载 作者:行者123 更新时间:2023-12-03 00:53:55 26 4
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编辑:我想我可能必须在 run() 函数中实现 BFS?

EDIT2:我已将代码更新为现在的代码。我对此不太满意,但它似乎成功地在电源导体上添加了星号。我必须输出房屋总数以及有多少房屋断电,但我不知道如何正确跟踪该数量。

EDIT3:好的,我完全相信我的网格填充是正确的。然而,我无法追踪没有通电的房屋数量。任何有关这方面的建议都会很棒。

这个想法是这样的:

给我一​​个充满符号的 h x w 矩阵。其中之一是字母“P”,代表发电厂。从该发电厂出发,任何大写字母都是导电的,可以将电力输送到 map 上。最重要的是,有 3 个符号代表电线也可以传输电力。它们是:+、- 和 |

最后,还有字母“H”代表的房屋。如果靠近任何其他导电符号,这些符号也具有导电性。这个想法是,如果电力可以到达,则用星号“*”填充网格。问题是每个发电厂 P 只能为 30 个家庭 H 供电。我发布的代码显然尚未完成,但广度优先填充算法(bff)已完成。我可以成功填充 P 旁边的任何直接符号,但我无法弄清楚如何继续使用不同的符号追踪 P。欢迎任何想法。

输入示例:

15 22
, , , , , , , , , , . . . , , , , , , , , ,
, , . , , , , , , H H H - - + , , , , , , ,
, , , , , , , , , H H H . . | . , , , . , ,
, , , , , , , , , H H H = , | . , , , . . .
, , , , , , , , , H H H = , | . , , , . . .
, , , . . . . . , H H H = , | , , , , . . .
. . . . . . . . , H H H = , | , , , , . . .
, , . , , = = = = = = = = , | , , , . . . .
, , X X X X , . . . . C C C C C C . . . . .
. . X P X X , . . . . C C C C C C . . . . .
. . X X X X - - - - - C C C C C C . . . . .
. ~ X X X X . . . . . . . . . . . . . . . .
~ ~ ~ ~ . . . . . . . . . . . . . . . . . .
~ ~ ~ ~ ~ . . . . . . . . . . . . . . . . ~
~ ~ ~ ~ ~ ~ . . . . . . . . . . . ~ ~ . ~ ~

示例输出:

0 of 18 homes are without power .
, , , , , , , , , , . . . , , , , , , , , ,
, , . , , , , , , * * * * * * , , , , , , ,
, , , , , , , , , * * * . . * . , , , . , ,
, , , , , , , , , * * * = , * . , , , . . .
, , , , , , , , , * * * = , * . , , , . . .
, , , . . . . . , * * * = , * , , , , . . .
. . . . . . . . , * * * = , * , , , , . . .
, , . , , = = = = = = = = , * , , , . . . .
, , * * * * , . . . . * * * * * * . . . . .
. . * * * * , . . . . * * * * * * . . . . .
. . * * * * * * * * * * * * * * * . . . . .
. ~ * * * * . . . . . . . . . . . . . . . .
~ ~ ~ ~ . . . . . . . . . . . . . . . . . .
~ ~ ~ ~ ~ . . . . . . . . . . . . . . . . ~
~ ~ ~ ~ ~ ~ . . . . . . . . . . . ~ ~ . ~ ~

到目前为止的代码:

public class PowerGrid {
String[][] grid = new String[120][120];
LinkedList<Pair> q = new LinkedList<Pair>();
LinkedList<Pair> visited = new LinkedList<Pair>();
Map<Pair, Integer> dist = new HashMap<Pair, Integer>();
ArrayList<Pair> pwrPlants = new ArrayList<Pair>();
String[] alphabet = {"A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L",
"M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z"};
int row = -1;
int col = -1;

int homesToPwr = 0;
int homesNotPwrd = 0;
int totalHomes = 0;

public static void main (String[] args) {
PowerGrid pg = new PowerGrid();
pg. run();

}

public void run() {
Scanner sc = new Scanner(System.in);

row = sc.nextInt();
col = sc.nextInt();

for (int i = 0; i < row; i++)
for (int j = 0; j < col; j++) {
grid[i][j] = sc.next();
}

for (int i = 0; i < row; i++)
for (int j = 0; j < col; j++) {
if (grid[i][j].equals("P")) {
Pair tmpPair = new Pair(i, j);
pwrPlants.add(tmpPair);
}
}

for (Pair p : pwrPlants) {
homesToPwr += 30;
for (String s : alphabet) {
if (s == "H")
if (homesToPwr != 0)
homesToPwr--;

bff(grid, p.x, p.y, s, "*");
}

bff(grid, p.x, p.y, "-", "*");
bff(grid, p.x, p.y, "+", "*");
bff(grid, p.x, p.y, "|", "*");
}

System.out.println(homesToPwr + " of " + totalHomes + " are without power.");

for (int i = 0; i < row; i++) {
for (int j = 0; j < col; j++) {
System.out.print(grid[i][j]);
}
System.out.println();
}
}

public void bff (String[][] grid, int x, int y, String oldSymbol, String newSymbol) {
if (oldSymbol == newSymbol) return;


Pair pair = new Pair(x,y);
q.addLast(pair);
dist.put(pair, 0);
grid[x][y] = newSymbol;

while (!q.isEmpty()) {
Pair v = q.getFirst();
q.pop();
int d = dist.get(v) + 1;
String[] symbols = {"+", "-", "|"};

visit(grid, v.x + 1, v.y, new Pair(v.x+1, v.y), oldSymbol, newSymbol, d);
visit(grid, v.x - 1, v.y, new Pair(v.x-1, v.y), oldSymbol, newSymbol, d);
visit(grid, v.x, v.y + 1, new Pair(v.x, v.y+1), oldSymbol, newSymbol, d);
visit(grid, v.x, v.y - 1, new Pair(v.x, v.y-1), oldSymbol, newSymbol, d);

for (String t : alphabet){
if (t == "H") {
if (homesToPwr != 0) {
homesToPwr--;
}
}

visit(grid, v.x + 1, v.y, new Pair(v.x+1, v.y), t, newSymbol, d);
visit(grid, v.x - 1, v.y, new Pair(v.x-1, v.y), t, newSymbol, d);
visit(grid, v.x, v.y + 1, new Pair(v.x, v.y+1), t, newSymbol, d);
visit(grid, v.x, v.y - 1, new Pair(v.x, v.y-1), t, newSymbol, d);

}
for (String s : symbols) {
visit(grid, v.x + 1, v.y, new Pair(v.x+1, v.y), s, newSymbol, d);
visit(grid, v.x - 1, v.y, new Pair(v.x-1, v.y), s, newSymbol, d);
visit(grid, v.x, v.y + 1, new Pair(v.x, v.y+1), s, newSymbol, d);
visit(grid, v.x, v.y - 1, new Pair(v.x, v.y-1), s, newSymbol, d);
}
}
}

public void visit (String[][] A, int x, int y, Pair pair, String oldSymbol, String newSymbol, int d) {
if ((x >= 0 && y >= 0) && (x < row && y < col) && A[x][y].equals(oldSymbol)) {
if (oldSymbol == "H")
totalHomes++;
A[x][y] = newSymbol;
dist.put(pair, d);
q.addLast(pair);

}
}

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

public String toString() {
String out = "X: " + this.x + " Y: " + this.y;
return out;
}
}
}

最佳答案

您的程序未正确计算房屋总数。

例如,这里有 19 栋房屋,但通过访问方法只统计到 18 栋。这是因为您只访问“连接”到发电厂的案例。

15 22
, , , , , , , , , , . . . , , , , , , , , ,
, , . , , , , H , H H H - - + , , , , , , ,
, , , , , , , , , H H H . . | . , , , . , ,
, , , , , , , , , H H H = , | . , , , . . .
, , , , , , , , , H H H = , | . , , , . . .
, , , . . . . . , H H H = , | , , , , . . .
. . . . . . . . , H H H = , | , , , , . . .
, , . , , = = = = = = = = , | , , , . . . .
, , X X X X , . . . . C C C C C C . . . . .
. . X P X X , . . . . C C C C C C . . . . .
. . X X X X - - - - - C C C C C C . . . . .
. ~ X X X X . . . . . . . . . . . . . . . .
~ ~ ~ ~ . . . . . . . . . . . . . . . . . .
~ ~ ~ ~ ~ . . . . . . . . . . . . . . . . ~
~ ~ ~ ~ ~ ~ . . . . . . . . . . . ~ ~ . ~ ~

您可以这样计算房屋总数:

    for (int i = 0; i < row; i++)
for (int j = 0; j < col; j++) {
grid[i][j] = sc.next();
if(grid[i][j].equals("H")){
totalHomes++;
}
}

然后用这种方法计算实际的供电房屋数量(我使用了一个新属性homesPwrd(=0),仅通过此方法修改,不确定您所有其他整数有什么用......):

public void visit (String[][] A, int x, int y, Pair pair, String oldSymbol, String newSymbol, int d) {
if ((x >= 0 && y >= 0) && (x < row && y < col) && A[x][y].equals(oldSymbol)) {
if (oldSymbol == "H")
homesPwrd++;
A[x][y] = newSymbol;
dist.put(pair, d);
q.addLast(pair);

}
}

终于

    System.out.println(totalHomes-homesPwrd + " of " + totalHomes + " are without power.");

输出

19 个中的 1 个断电。

    1 of 19 are without power.
,,,,,,,,,,...,,,,,,,,,
,,.,,,,H,******,,,,,,,
,,,,,,,,,***..*.,,,.,,
,,,,,,,,,***=,*.,,,...
,,,,,,,,,***=,*.,,,...
,,,.....,***=,*,,,,...
........,***=,*,,,,...
,,.,,========,*,,,....
,,****,....******.....
..****,....******.....
..***************.....
.~****................
~~~~..................
~~~~~................~
~~~~~~...........~~.~~

关于java - 使用广度优先填充算法填充 "power grid";我被困住了,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/16259519/

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