1. 拓扑排序问题(HDU 1285)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.StreamTokenizer;public class Main
{static int[] inDegree = new int[502];static int[][] v = new int[502][502];static boolean[] contain = new boolean[502];public static void main(String[] args) throws IOException{StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));int n, m;while (in.nextToken() != StreamTokenizer.TT_EOF){n = (int) in.nval;in.nextToken();m = (int) in.nval;init(n);for (int i = 1; i <= m; i++){in.nextToken();int p1 = (int) in.nval;in.nextToken();int p2 = (int) in.nval;if(v[p1][p2] == 0){inDegree[p2]++;}v[p1][p2] = 1;}topologicalOrder(n);System.out.println();}}private static void topologicalOrder(int n){// TODO Auto-generated method stubint vv;for (int i = 1; i <= n; i++){for (int j = 1; j <= n; j++){if (inDegree[j] == 0 && contain[j]){vv = j;contain[vv] = false;for (int j2 = 1; j2 <= n; j2++){if (v[vv][j2] > 0){inDegree[j2]--;}}System.out.print(vv);if(i != n){System.out.print(" ");}break;}}}}private static void init(int n){// TODO Auto-generated method stubfor (int i = 1; i <= n; i++){for (int j = 1; j <= n; j++){v[i][j] = 0;}inDegree[i] = 0;contain[i] = true;}}}

上述代码就是最简单的拓扑排序，里面有很多可以优化的地方，比如用如果数据稀疏的话可以用邻接表来代替二维数组（矩阵），在找编号最小的且度为0的顶点，可以使用最小堆（ 优先队列）。下面代码使用优先队列来代替原本的遍历查找，使复杂度降低。

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.StreamTokenizer;
import java.util.PriorityQueue;
import java.util.Queue;public class Main
{static int[] inDegree = new int[502];static int[][] v = new int[502][502];public static void main(String[] args) throws IOException{StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));int n, m;while (in.nextToken() != StreamTokenizer.TT_EOF){n = (int) in.nval;in.nextToken();m = (int) in.nval;init(n);for (int i = 1; i <= m; i++){in.nextToken();int p1 = (int) in.nval;in.nextToken();int p2 = (int) in.nval;if (v[p1][p2] == 0){inDegree[p2]++;}v[p1][p2] = 1;}topologicalOrder(n);System.out.println();}}private static void topologicalOrder(int n){// TODO Auto-generated method stubint vv;Queue<Integer> queue = new PriorityQueue<Integer>();for (int i = 1; i <= n; i++){if (inDegree[i] == 0){queue.add(i);}}for (int i = 1; i <= n; i++){vv = queue.poll();for (int j2 = 1; j2 <= n; j2++){if (v[vv][j2] > 0){inDegree[j2]--;if (inDegree[j2] == 0){queue.add(j2);}}}System.out.print(vv);if (i != n){System.out.print(" ");}}}private static void init(int n){// TODO Auto-generated method stubfor (int i = 1; i <= n; i++){for (int j = 1; j <= n; j++){v[i][j] = 0;}inDegree[i] = 0;}}}

直接把入度为0的顶点放到优先队列中（普通的拓扑排序只用普通队列/栈/数组即可，这里题目中有要求，相同入度为0的点要求序号小的先输出，所以使用优先队列），使用队列减少了一遍遍的遍历所有点的入度。

在使用BFS和Dijkstra的优先队列中都要使用到队列，关于Dijkstra以及最短路径算法，请查看最短路径算法

系列：

【算法系列 一】 Linked List

【算法系列 二】 Stack

【算法系列 三】 Quene

【算法系列 四】 String