HDU 1142 A Walk Through the Forest

A Walk Through the Forest

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3436    Accepted Submission(s): 1261

Problem Description

        Jimmy experiences a lot of stress at work these days, especially since his accident made working difficult. To relax after a hard day, he likes to walk home. To make things even nicer, his office is on one side of a forest, and his house is on the other. A nice walk through the forest, seeing the birds and chipmunks is quite enjoyable.
The forest is beautiful, and Jimmy wants to take a different route everyday. He also wants to get home before dark, so he always takes a path to make progress towards his house. He considers taking a path from A to B to be progress if there exists a route from B to his home that is shorter than any possible route from A. Calculate how many different routes through the forest Jimmy might take.

 

Input

        Input contains several test cases followed by a line containing 0. Jimmy has numbered each intersection or joining of paths starting with 1. His office is numbered 1, and his house is numbered 2. The first line of each test case gives the number of intersections N, 1 < N ≤ 1000, and the number of paths M. The following M lines each contain a pair of intersections a b and an integer distance 1 ≤ d ≤ 1000000 indicating a path of length d between intersection a and a different intersection b. Jimmy may walk a path any direction he chooses. There is at most one path between any pair of intersections.

 

Output

        For each test case, output a single integer indicating the number of different routes through the forest. You may assume that this number does not exceed 2147483647

 

Sample Input

5 61 3 21 4 23 4 31 5 124 2 345 2 247 81 3 11 4 13 7 17 4 17 5 16 7 15 2 16 2 10

 

Sample Output

24
package hdu;import java.util.ArrayDeque;import java.util.ArrayList;import java.util.Arrays;import java.util.Queue;import java.util.Scanner;/** * @description HDU 1142 A Walk Through the Forest * @technique最短路径，动态规划。 * @date20120821 * @time14:24 * @version1.1 * @author Alex */public class Hdu1142_20120821_0 {public static void main(String[] args) {Scanner in = new Scanner(System.in);int n,m,temp,x,y,i;Queue<Integer> queue = new ArrayDeque<Integer>();int [][] map;int [] dis;int [] dp;while(0!=(n = in.nextInt())){m = in.nextInt();map = new int[n][n];dis = new int[n];Arrays.fill(dis, Integer.MAX_VALUE);dis[1] = 0;for(i = 0; i < m; ++i){x = in.nextInt() - 1;y = in.nextInt() - 1;temp = in.nextInt();map[x][y] = temp;map[y][x] = temp;}queue.add(1);bellmanFord(map,queue,dis);//buildTree(map,dis);dp = new int[dis.length];Arrays.fill(dp, -1);dp[1] = 1;int count = trave(map,0,dp,dis);System.out.println(count);queue.clear();}}private static void bellmanFord(int [][] map, Queue<Integer> queue, int [] dis){while(!queue.isEmpty()){int i,j;i = queue.poll();for(j = 0; j < map.length; ++j){if(map[i][j] + dis[i] < dis[j] && map[i][j] != 0){dis[j] = map[i][j] + dis[i];if(!queue.contains(j)){queue.add(j);}}}}}/** * 跟据最短路径，建立路径树。树叶子个数就是路径数。 * @param map * @param dis */@SuppressWarnings("unused")private static void buildTree(int [][] map, int[] dis){int i,j;Node [] tree = new Node[dis.length];for(i = 0; i < dis.length; ++i){tree[i] = newNode(i);}Queue<Integer> queue = new ArrayDeque<Integer>();queue.add(0);while(!queue.isEmpty()){i = queue.poll();for(j = 0; j < dis.length; ++j){if(!tree[i].next.contains(j)&&map[i][j] != 0 && dis[i] > dis[j]){tree[i].next.add(j);queue.add(j);}}}int [] dp = new int[dis.length];Arrays.fill(dp, -1);dp[1] = 1;int count = trave(tree,0,dp);  //遍历路径树。返回树叶子个数。System.out.println(count); //输出结果。}/** * 查找路径时使用动态规划的思想。不然会超时。 * @param tree * @param i * @param dp * @return　路径数。 */private static int trave(Node [] tree, int i, int [] dp){if(dp[i] != -1){return dp[i];}dp[i] = 0;for(int j = 0; j < tree[i].next.size(); ++j){dp[i] += trave(tree,tree[i].next.get(j),dp);}return dp[i];}private static int trave(int [][] map,int i, int [] dp, int [] dis){if(dp[i] != -1){return dp[i];}if(dp[i] == 1){return dp[i];}dp[i] = 0;for(int j = 0; j < map.length; ++j){if(dis[i] > dis[j] && map[i][j] != 0)dp[i] += trave(map,j,dp,dis);}return dp[i];}private static Node newNode(int i){return new Hdu1142_20120821_0().new Node(i);}private class Node{int i;ArrayList<Integer> next;public Node(int i){this.i = i;this.next = new ArrayList<Integer>();}}}

 

PS： 这个代码是上一次那个代码的改写，上次代码建了一个路径树。这次用的是记忆化搜索。两次代都在这里，可以对比看，现以这个版本比上个版本在时间上快了 1000ms