在我的作業(yè)中，有一個(gè)術(shù)語(yǔ)，在兩個(gè)節(jié)點(diǎn)之間，先前單擊的節(jié)點(diǎn)應(yīng)保留在屏幕上，并且應(yīng)該有線條將每個(gè)節(jié)點(diǎn)與下一個(gè)節(jié)點(diǎn)連接起來(lái)（最后一個(gè)節(jié)點(diǎn)連接到第一個(gè)節(jié)點(diǎn)）。除了這些線（當(dāng)然還有網(wǎng)格線）之外不應(yīng)該有其他線。我嘗試通過(guò)在數(shù)組內(nèi)使用循環(huán)來(lái)做到這一點(diǎn)。但這不是我正在尋找的答案......int [] a;int [] b;ArrayList<Point> points = new ArrayList<Point>();class Point { int x, y, r; Point(int x, int y, int r) { this.x = x; this.y = y; this.r = r; } void Draw() { circle(this.x, this.y, this.r*2); }}int n_part=10;void setup(){ size(600,360); a = new int [100]; b = new int [100];}void draw () { background (255); int gridW = width/n_part; int gridH = height/n_part; stroke(210); noFill(); for (int row = 0; row < n_part; row++){ int gridY = 0 + row*gridH; for (int col = 0; col < n_part; col++) { int gridX = 0+ col* gridW; rect (gridX, gridY, gridW, gridH); } } stroke(0, 0, 0); fill(0); for (int i = 0; i < points.size(); ++ i) { Point p = points.get(i); p.Draw(); }}void mousePressed() { int gridW = width/n_part; int gridH = height/n_part; int x = round(mouseX / (float)gridW) * gridW; int y = round(mouseY / (float)gridH) * gridH; points.add(new Point(x, y, 5)); { for (int j=0; j < a.length; j++) a [j] = (mouseX / (int)gridW) * gridW; for (int k=0; k < b.length; k++) b [k] = height; }}void mouseReleased(){ for (int k=0; k < a.length; k++) line (0, 0, a [k], b [k]);}
1 回答
狐的傳說(shuō)
TA貢獻(xiàn)1804條經(jīng)驗(yàn) 獲得超3個(gè)贊
您所要做的就是在連續(xù)點(diǎn)之間畫(huà)線。
點(diǎn)的下一個(gè)索引i是i+1。%對(duì)于最后一個(gè)點(diǎn)，它是 0。這可以通過(guò)模 ( ) 運(yùn)算符 (h )來(lái)實(shí)現(xiàn)(i+1) % points.size()。
獲取循環(huán)中的 2 個(gè)連續(xù)點(diǎn)并在它們之間畫(huà)一條線：
for (int i = 0; i < points.size(); ++ i) {
? ? Point p1 = points.get(i);
? ? Point p2 = points.get((i+1) % points.size());
? ? line(p1.x, p1.y, p2.x, p2.y);? ?
}
參見(jiàn)示例：
void draw () {
? ? background (255);
? ? int gridW = width/n_part;
? ? int gridH = height/n_part;
? ? strokeWeight(1);
? ? stroke(210);
? ? noFill();
? ? for (int row = 0; row < n_part; row++){?
? ? ? ? int gridY = 0 + row*gridH;
? ? ? ? for (int col = 0; col < n_part; col++) {
? ? ? ? ? ? int gridX = 0+ col* gridW;
? ? ? ? ? ? rect (gridX, gridY, gridW, gridH);? ?
? ? ? ? }
? ? }
? ? stroke(0, 0, 0);
? ? fill(0);
? ? for (int i = 0; i < points.size(); ++ i) {
? ? ? ? Point p = points.get(i);
? ? ? ? p.Draw();
? ? }
? ? strokeWeight(3);
? ? stroke(255, 0, 0);
? ? for (int i = 0; i < points.size(); ++ i) {
? ? ? ? Point p1 = points.get(i);
? ? ? ? Point p2 = points.get((i+1) % points.size());
? ? ? ? line(p1.x, p1.y, p2.x, p2.y);? ?
? ? }
}
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