1310 One-way traffic

In a certain town there are n intersections connected by two- and one-way streets. The town is very
modern so a lot of streets run through tunnels or viaducts. Of course it is possible to travel between
any two intersections in both ways, i.e. it is possible to travel from an intersection a to an intersection
b as well as from b to a without violating traffic rules. Because one-way streets are safer, it has been
decided to create as much one-way traffic as possible. In order not to make too much confusion it has
also been decided that the direction of traffic in already existing one-way streets should not be changed.
Your job is to create a new traffic system in the town. You have to determine the direction of
traffic for as many two-way streets as possible and make sure that it is still possible to travel both ways
between any two intersections.
Write a program that:
? reads a description of the street system in the town from the standard input,
? for each two-way street determines one direction of traffic or decides that the street must remain
two-way,
? writes the answer to the standard output.

Input
The first line of the input contains two integers n and m, where 2 ≤ n ≤ 2000 and n?1 ≤ m ≤ n(n?1)/2.
Integer n is the number of intersections in the town and integer m is the number of streets.
Each of the next m lines contains three integers a, b and c, where 1 ≤ a ≤ n, 1 ≤ b ≤ n, a ?= b and
c belongs to {1, 2}. If c = 1 then intersections a and b are connected by an one-way street from a to
b. If c = 2 then intersections a and b are connected by a two-way street. There is at most one street
connecting any two intersections.

Output
The output contains exactly the same number of lines as the number of two-way streets in the input.
For each such street (in any order) the program should write three integers a, b and c meaning, the new
direction of the street from a to b (c = 1) or that the street connecting a and b remains two-way (c = 2).
If there are more than one solution with maximal number of one-way streets then your program should
output any of them but just one.

Sample Input
4 4
4 1 1
4 2 2
1 2 1
1 3 2


Sample Output
2 4 1
3 1 2

package com.liuzhen.practice;

import java.util.ArrayList;
import java.util.Scanner;
import java.util.Stack;

public class Main {
    public static int n;  //给定图的顶点数
    public static int count;  //记录遍历次序
    public static int[] DFN;
    public static int[] Low;
    public static int[] parent;   //parent[i] = j，表示顶点i的直接父母顶点为j
    public static Stack<Integer> stack;
    public static ArrayList<edge>[] map;
    public static ArrayList<edge> ans;  //存储最终输出结果
    
    static class edge {
        public int a;  //边的起点
        public int b;  //边的终点
        public int c;  //c = 1表示单向边，c = 2表示双向边
        
        public edge(int a, int b, int c) {
            this.a = a;
            this.b = b;
            this.c = c;
        }
    }
    
    @SuppressWarnings("unchecked")
    public void init() {
        count = 0;
        DFN = new int[n + 1];
        Low = new int[n + 1];
        parent = new int[n + 1];
        stack = new Stack<Integer>();
        map = new ArrayList[n + 1];
        ans = new ArrayList<edge>();
        for(int i = 1;i <= n;i++) {
            DFN[i] = -1;
            Low[i] = -1;
            parent[i] = -1;
            map[i] = new ArrayList<edge>();
        }
    }
    
    public void TarJan(int start, int father) {
        DFN[start] = count++;
        Low[start] = DFN[start];
        parent[start] = father;
        stack.push(start);
        for(int i = 0;i < map[start].size();i++) {
            edge temp = map[start].get(i);
            int j = temp.b;
            if(DFN[j] == -1) {
                TarJan(j, start);
                Low[start] = Math.min(Low[start], Low[j]);
                if(temp.c == 2) {
                    if(Low[j] > DFN[start]) {   //当边temp为割边(或者桥)时
                        ans.add(temp);
                    } else {
                        ans.add(new edge(temp.a, temp.b, 1));
                    }
                }
            } else if(j != parent[start]) {  //当j不是start的直接父母节点时
                Low[start] = Math.min(Low[start], DFN[j]);
                if(temp.c == 2) {
                    ans.add(new edge(temp.a, temp.b, 1));
                }
            }
        }
    }
    
    public void getResult() {
        for(int i = 1;i <= n;i++) {
            if(parent[i] == -1)
                TarJan(i, 0);
        }
        for(int i = 0;i < ans.size();i++)
            System.out.println(ans.get(i).a+" "+ans.get(i).b+" "+ans.get(i).c);
    }
    
    public static void main(String[] args) {
        Main test = new Main();
        Scanner in = new Scanner(System.in);
        n = in.nextInt();
        int k = in.nextInt();
        test.init();
        for(int i = 0;i < k;i++) {
            int a = in.nextInt();
            int b = in.nextInt();
            int c = in.nextInt();
            map[a].add(new edge(a, b, c));
            if(c == 2)
                map[b].add(new edge(b, a, c));
        }
        test.getResult();
    }
}

4 4
4 1 1
4 2 2
1 2 1
1 3 2
2 4 1
1 3 2