poj 2051 最小堆

悲剧的poj2051 最小堆，奉献了无数次后才AC。这道题就是个最小堆，把最先到达的query输出query编号，然后继续排队等待同样的时间，要是某几个请求同时到达的话，就要按照编号从小到大的输出。通过这道题感觉自己考虑问题不够全面。

废话少说了，看下面的代码吧：

package poj; import java.util.Scanner; public class Main2051 { private static Query[] query = new Query[1002]; public static void main(String[] args) { Scanner sc = new Scanner(System.in); int i = 0; while (sc.hasNext()) { String s = sc.nextLine(); if (s.equals("#")) { int k = sc.nextInt(); slove(k, i); break; } String[] str = s.split(" "); Query q = new Query(); q.d = Integer.parseInt(str[1].trim()); q.w = Integer.parseInt(str[2].trim()); q.p = q.w; insertHeap(q, i); i++; } } public static void slove(int k, int n) { // System.out.println("start........."); while (k > 0) { System.out.println(query[0].d); query[0].w += query[0].p; // System.out.println(query[0].w); keapHeap(n); --k; } } public static void keapHeap(int n) { Query q = query[0]; // System.out.println(q.w + ";" + q.d); int i = 0; int j = (i << 1) + 1; while (j <= n - 1) { if (j < n - 1 && query[j].w > query[j + 1].w) j++; else if (j < n - 1 && query[j].w == query[j + 1].w && query[j].d > query[j + 1].d) { j++; } if (query[j].w < q.w) { query[i] = query[j]; i = j; } else if (query[j].w > q.w) break; else if (query[j].w == q.w && query[j].d < q.d) { query[i] = query[j]; i = j; } else break; j = (i << 1) + 1; } query[i] = q; } public static void insertHeap(Query q, int n) { // System.out.println("n = " + n); if (n == 0) { query[0] = q; } else { int i = n; int j = (i - 1) >> 1; while (j >= 0) { if (query[j].w > q.w) { query[i] = query[j]; i = j; } else if (query[j].w == q.w && query[j].d > q.d) { query[i] = query[j]; i = j; } else break; j = (i - 1) >> 1; } query[i] = q; } } static class Query { private int w; private int p; private int d; } }