检测二维数组中的相干邻居/邻域
Detect coherent neighbors / neighborhood in 2d array
我有一个任意的二维数组,每个字段都有一个 id 和一个 teamid(这里用颜色 1 表示)。
我希望每个社区都有一个包含 id 的数组
在里面。
一个社区由具有相同 teamid 水平和垂直(不是对角线)邻居的字段组成
例如:
这就是我所拥有的:
array[0][0] = {id:1,teamId:1}
array[1][0] = {id:2,teamId:1}
array[2][0] = {id:3,teamId:0}
array[3][0] = {id:4,teamId:2}
array[4][0] = {id:5,teamId:2}
array[5][0] = {id:6,teamId:0}
array[0][1] = {id:7,teamId:1}
array[1][1] = {id:8,teamId:1}
array[2][1] = {id:9,teamId:1}
array[3][1] = {id:10,teamId:2}
array[4][1] = {id:11,teamId:2}
array[5][1] = {id:12,teamId:0}
//and so on..
这就是我想要的:
neighborhood[1] = [1,2,7,8,9,13,14]
neighborhood[2] = [4,5,10,11]
neighborhood[3] = [16,22,23,24,29,30]
neighborhood[4] = [25,31,32,37,38]
neighborhood[5] = [35,41]
我不是在搜索图像,而是在搜索数组
neighborhood
提前致谢!
您可以使用点和方块游戏中的逻辑。如果玩家用墙将其包围,则该块属于该玩家。因此,除了外部单元之外,每个单元还需要 4 面墙。要测试单元格是否已关闭,您可以使用 4 class 个变量:
var Block = function() {
this.isclosed=0;
this.left=0;
this.top=0;
this.right=0;
this.bottom=0;
return this;
}
Block.prototype = {
isClosed : function () {
if (this.isclosed==true) {
return false;
} else if (this.left && this.top && this.right && this.bottom) {
this.isclosed=true;
return true;
} else {
return this.left && this.top && this.right && this.bottom;
}
}
}
您可以试试我的点和块游戏实现 @ https://dotsgame.codeplex.com/。
解决该问题的方法参考Connected Component Labelling
有人问过一次similar question,我从中得到了解决方案:
// matrix dimensions
var row_count = 20;
var col_count = 20;
var numOfTeams = 2;
// the input matrix
var m = [];
// the labels, 0 means unlabeled
var label = [];
var source = document.getElementById("source");
for (var i = 0; i < row_count; i++) {
var row = source.insertRow(0);
m[i] = [];
label[i] = [];
for (var j = 0; j < col_count; j++) {
//m[i][j] = Math.round(Math.random());
m[i][j] = getRandomInt(0, numOfTeams + 1);
label[i][j] = 0;
var cell1 = row.insertCell(0);
cell1.innerHTML = m[i][j];
}
}
// direction vectors
var dx = [1, 0, -1, 0];
var dy = [0, 1, 0, -1];
function dfs(x, y, current_label, team) {
if (x < 0 || x == row_count) return; // out of bounds
if (y < 0 || y == col_count) return; // out of bounds
if (label[x][y] || team != m[x][y]) return; // already labeled or not marked with 1 in m
// mark the current cell
label[x][y] = current_label;
// recursively mark the neighbors
for (var direction = 0; direction < 4; ++direction) {
dfs(x + dx[direction], y + dy[direction], current_label, team);
}
}
function find_components() {
var component = 0;
for (var i = 0; i < row_count; ++i) {
for (var j = 0; j < col_count; ++j) {
if (!label[i][j] && m[i][j]) dfs(i, j, ++component, m[i][j]);
}
}
}
find_components();
var result = document.getElementById("result");
for (var i in label) {
var string = ""
var row = result.insertRow(0);
for (var j in label[i]) {
string += label[i][j] + " "
var cell1 = row.insertCell(0);
cell1.innerHTML = label[i][j];
}
}
function getRandomInt(min, max) {
return Math.floor(Math.random() * (max - min)) + min;
}
table tr td {
min-width: 14px
}
<div style="float:left">
<table id="source"></table>
</div>
<div style="float:right">
<table id="result"></table>
</div>
我有一个任意的二维数组,每个字段都有一个 id 和一个 teamid(这里用颜色 1 表示)。
我希望每个社区都有一个包含 id 的数组 在里面。
一个社区由具有相同 teamid 水平和垂直(不是对角线)邻居的字段组成
例如: 这就是我所拥有的:
array[0][0] = {id:1,teamId:1}
array[1][0] = {id:2,teamId:1}
array[2][0] = {id:3,teamId:0}
array[3][0] = {id:4,teamId:2}
array[4][0] = {id:5,teamId:2}
array[5][0] = {id:6,teamId:0}
array[0][1] = {id:7,teamId:1}
array[1][1] = {id:8,teamId:1}
array[2][1] = {id:9,teamId:1}
array[3][1] = {id:10,teamId:2}
array[4][1] = {id:11,teamId:2}
array[5][1] = {id:12,teamId:0}
//and so on..
这就是我想要的:
neighborhood[1] = [1,2,7,8,9,13,14]
neighborhood[2] = [4,5,10,11]
neighborhood[3] = [16,22,23,24,29,30]
neighborhood[4] = [25,31,32,37,38]
neighborhood[5] = [35,41]
我不是在搜索图像,而是在搜索数组
neighborhood
提前致谢!
您可以使用点和方块游戏中的逻辑。如果玩家用墙将其包围,则该块属于该玩家。因此,除了外部单元之外,每个单元还需要 4 面墙。要测试单元格是否已关闭,您可以使用 4 class 个变量:
var Block = function() {
this.isclosed=0;
this.left=0;
this.top=0;
this.right=0;
this.bottom=0;
return this;
}
Block.prototype = {
isClosed : function () {
if (this.isclosed==true) {
return false;
} else if (this.left && this.top && this.right && this.bottom) {
this.isclosed=true;
return true;
} else {
return this.left && this.top && this.right && this.bottom;
}
}
}
您可以试试我的点和块游戏实现 @ https://dotsgame.codeplex.com/。
解决该问题的方法参考Connected Component Labelling
有人问过一次similar question,我从中得到了解决方案:
// matrix dimensions
var row_count = 20;
var col_count = 20;
var numOfTeams = 2;
// the input matrix
var m = [];
// the labels, 0 means unlabeled
var label = [];
var source = document.getElementById("source");
for (var i = 0; i < row_count; i++) {
var row = source.insertRow(0);
m[i] = [];
label[i] = [];
for (var j = 0; j < col_count; j++) {
//m[i][j] = Math.round(Math.random());
m[i][j] = getRandomInt(0, numOfTeams + 1);
label[i][j] = 0;
var cell1 = row.insertCell(0);
cell1.innerHTML = m[i][j];
}
}
// direction vectors
var dx = [1, 0, -1, 0];
var dy = [0, 1, 0, -1];
function dfs(x, y, current_label, team) {
if (x < 0 || x == row_count) return; // out of bounds
if (y < 0 || y == col_count) return; // out of bounds
if (label[x][y] || team != m[x][y]) return; // already labeled or not marked with 1 in m
// mark the current cell
label[x][y] = current_label;
// recursively mark the neighbors
for (var direction = 0; direction < 4; ++direction) {
dfs(x + dx[direction], y + dy[direction], current_label, team);
}
}
function find_components() {
var component = 0;
for (var i = 0; i < row_count; ++i) {
for (var j = 0; j < col_count; ++j) {
if (!label[i][j] && m[i][j]) dfs(i, j, ++component, m[i][j]);
}
}
}
find_components();
var result = document.getElementById("result");
for (var i in label) {
var string = ""
var row = result.insertRow(0);
for (var j in label[i]) {
string += label[i][j] + " "
var cell1 = row.insertCell(0);
cell1.innerHTML = label[i][j];
}
}
function getRandomInt(min, max) {
return Math.floor(Math.random() * (max - min)) + min;
}
table tr td {
min-width: 14px
}
<div style="float:left">
<table id="source"></table>
</div>
<div style="float:right">
<table id="result"></table>
</div>