PTT數位生活區 / Prob_Solve (計算數學 Problem Solving)

[問題] Google Code Jam 2009, Round 3 - C

看板Prob_Solve (計算數學 Problem Solving)作者RockLee (Now of all times)時間12年前 (2012/03/15 18:26)推噓0(0推 0噓 0→)

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題目敘述網址: http://code.google.com/codejam/contest/243103/dashboard#s=p2 沒有想出來如何判斷是否為 3-colorable, 看了當初參賽者上傳的 code 還是不太懂 ~><~ Invader 的 code 中用來判斷是否為 3-colorable 主要為以下幾行: int count = 0; int u = pru[f1]; while (u != pru[f2]) {u = pr[u]; count++;} int d = prd[f1]; while (d != prd[f2]) {d = pr[d]; count++;} if (count % 2 == 1) {result = 4; break;} 看起來的意思似乎是, 若同一列的兩個相鄰球員間, 前一列在這兩個相鄰球員間的人數與後一列在這兩個相鄰球員間的人數和為奇數, 則非 3-colorable. 但我不懂的是為何這樣就可確認非 3-colorable, 例如以下三列(數字代表顏色): _1_2_1_ 2_____0 __1_2__ 看起來似乎滿足前後列人數和為奇數, 但仍為 3-colorable, 小弟是否哪邊誤解了呢? 麻煩版上高手開示一下 3-colorable 如何判斷 <_.._> [Invader's solution in Java] import java.io.*; import java.util.*; public class C implements Runnable { private String IFILE = "C-large-practice.in"; private Scanner in; private PrintWriter out; private class Pset implements Comparable<Pset> { int x; int y; public Pset(int x, int y) { this.x = x; this.y = y; } public int compareTo(Pset o) { return o.x - x; } @Override public String toString() { return "(" + x + ", " + y + ")"; } } public void Run() throws IOException { in = new Scanner(new File(IFILE)); out = new PrintWriter("output.txt"); int ntest = in.nextInt(); for (int test = 1; test <= ntest; test++) { int n = in.nextInt(); Pset[] mas = new Pset[n]; for(int i = 0; i < n; i++) {mas[i] = new Pset(in.nextInt(), in.nextInt());} Arrays.sort(mas); int[] pru = new int[n]; int[] pr = new int[n]; int[] prd = new int[n]; Arrays.fill(pru, -1); Arrays.fill(pr, -1); Arrays.fill(prd, -1); int result = 1; for (int i = 0; i < n; i++) { for(int j = i - 1; j >= 0; j--) if (mas[i].y - 1 == mas[j].y) { pru[i] = j; result = 2; break; } for(int j = i - 1; j >= 0; j--) if (mas[i].y == mas[j].y) { pr[i] = j; result = 2; break; } for(int j = i - 1; j >= 0; j--) if (mas[i].y + 1== mas[j].y) { prd[i] = j; result = 2; break; } } for (int i = 0; i < n; i++) if (pr[i] >= 0) { int f1 = i; int f2 = pr[f1]; if (pru[f1] >= 0 && pru[f2] >= 0 && pru[f1] == pru[f2] || prd[f1] >= 0 && prd[f2] >= 0 && prd[f1] == prd[f2] || pru[f1] >= 0 && prd[pru[f1]] == f2 || prd[f1] >= 0 && pru[prd[f1]] == f2) { result = 3; break; } } for(int i = 0; i < n; i++) if (pr[i] >= 0 && pr[pr[i]] >= 0) { int f1 = i; int f2 = pr[f1]; int f3 = pr[f2]; if (pru[f1] == -1 || prd[f1] == -1) continue; if (pru[f2] == -1 || prd[f2] == -1) continue; int count = 0; int u = pru[f1]; while (u != pru[f2]) { u = pr[u]; count++; } int d = prd[f1]; while (d != prd[f2]) { d = pr[d]; count++; } if (count % 2 == 1) { result = 4; break; } } out.println("Case #" + test + ": " + result); } in.close(); out.close(); } public void run() { try {Run();} catch(IOException e) {} } public static void main(String[] args) throws IOException { new C().Run(); } } -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 59.126.30.95