poj 1703 Find them, Catch them

A Walk Through the Forest

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3450    Accepted Submission(s): 1269

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
 
 
import java.io.BufferedReader;import java.io.IOException;import java.io.InputStreamReader;import java.io.StreamTokenizer;/** *  @功能Function Description:    并查集 *  @开发环境Environment:         eclipse * @技术特点Technique:           看题，知道是并查集，但是怎么写呢？并查集要得是a与b相同才合并，现在是a与b不同。  关键就是一句话：敌人的敌人是朋友。                                  我们对每个人x，都假设初始时他有一个敌人x+n，每次碰到D a b，就把a与b的敌人合并，b与a的敌人合并。                                  当查询时：若find(a)== find(b)：朋友                                  若find(a)==find(b+n)||find(b)==find(a+n)：敌人；                                  都不是：无法确定。 *@版本Version:   *@日期Date:                    20120815 *@备注Notes:                    */public class PK1703_3_20120815 {public static int[] father = new int[200005];public static void main(String[] args) throws IOException {StreamTokenizer st = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));st.nextToken();int countCase = (int)st.nval;int n, m;int x, y;String command;while(countCase-- !=0) {st.nextToken();n = (int)st.nval;//n个人st.nextToken();m = (int)st.nval;//m个消息序列makeSet(2*n);for(int i=0; i<m; i++) {st.nextToken();command = st.sval;//那种指令st.nextToken();x = (int)st.nval; st.nextToken();y = (int)st.nval;if("D".equals(command)) {union(x, y+n);union(y, x+n);} else {if(n==2) {if(x==y) System.out.println("In the same gang.");else System.out.println("In different gangs.");} else {System.out.println(judge(x, y, n));}}}}}public static String judge(int x, int y, int n) {int fx = find(x);int fy = find(y);if(fx == fy) {//如果根相等的话，证明输入过(D x ~)和(D y ~) return "In the same gang.";} else {if(fx==find(y+n) || fy==find(x+n)) {//输入过(x,y)，或者(y,x),因此x和y不同帮派return "In different gangs.";} else {return "Not sure yet.";}}}public static void union(int x, int y) {x = find(x);y = find(y);if(x != y) {father[x] = y;}}public static void makeSet(int n) {for(int i=1; i<=n; i++) {father[i] = i;}}public static int find(int x) {if(x != father[x]) {father[x] = find(father[x]);}return father[x];}}