Walking the Approximate Longest Path
Problem Statement :
Jenna is playing a computer game involving a large map with n cities numbered sequentially from 1 to n that are connected by m bidirectional roads. The game's objective is to travel to as many cities as possible without visiting any city more than once. The more cities the player visits, the more points they earn. As Jenna's fellow student at Hackerland University, she asks you for help choosing an optimal path. Given the map, can you help her find a path that maximizes her score? Note: She can start and end her path at any two distinct cities. Input Format The first line contains two space-separated integers describing the respective values of n (the number of cities) and m (the number of roads). Each line i of the m subsequent lines contains two space-separated integers, xi and yi, describing a bidirectional road between cities xi and yi. Constraints 1 <= n <= 10^4' 1 <= m <= 10^5 1 <= xi, yi <= n Output Format Print the following two lines of output: The first line must contain a single integer, d, denoting the length of the path. The second line must contain d distinct space-separated integers describing Jenna's path in the same order in which she visited each city
Solution :
Solution in C :
In C++ :
#include <cstdlib>
#include <cstdio>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <cstring>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
#define SIZE(x) (int((x).size()))
#define rep(i,l,r) for (int i=(l); i<=(r); i++)
#define repd(i,r,l) for (int i=(r); i>=(l); i--)
#define rept(i,c) for (typeof((c).begin()) i=(c).begin(); i!=(c).end(); i++)
#ifndef ONLINE_JUDGE
#define debug(x) { cerr<<#x<<" = "<<(x)<<endl; }
#else
#define debug(x) {}
#endif
#define maxn 10010
vector<int> e[maxn];
int vis[maxn], seq[2*maxn], ans[2*maxn];
void lemon()
{
int n,m; scanf("%d%d",&n,&m);
rep(i,1,m)
{
int x,y; scanf("%d%d",&x,&y);
e[x].push_back(y); e[y].push_back(x);
}
int maxv=0;
rep(iter,1,10)
{
memset(vis,0,sizeof vis);
seq[1+maxn]=rand()%n+1; vis[seq[1+maxn]]=1;
int beg = 1, all=1, last=seq[1+maxn], noprogress=0;
while (noprogress<200)
{
int k=rand()%e[last].size();
k=e[last][k];
if (!vis[k])
{
vis[k]=1; all++; seq[all+maxn]=k; last=k;
noprogress=0;
}
else
{
int where;
rep(i,beg,all) if (seq[i+maxn]==k) { where=i; break; }
reverse(seq+where+maxn+1,seq+all+maxn+1);
last=seq[all+maxn];
noprogress++;
}
}
last=seq[1+maxn], noprogress=0;
while (noprogress<200 && all-beg+1<n)
{
int k=rand()%e[last].size();
k=e[last][k];
if (!vis[k])
{
vis[k]=1; beg--; seq[beg+maxn]=k; last=k;
noprogress=0;
}
else
{
int where;
rep(i,beg,all) if (seq[i+maxn]==k) { where=i; break; }
reverse(seq+beg+maxn,seq+where+maxn);
last=seq[beg+maxn];
noprogress++;
}
}
if (all-beg+1>maxv)
{
maxv=all-beg+1; int c=0;
rep(i,beg,all)
{
c++; ans[c]=seq[i+maxn];
}
}
}
printf("%d\n",maxv);
rep(i,1,maxv) printf("%d ",ans[i]); printf("\n");
}
int main()
{
srand(time(0));
ios::sync_with_stdio(true);
#ifndef ONLINE_JUDGE
// freopen("5.in","r",stdin);
#endif
lemon();
return 0;
}
In Java ;
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int m = in.nextInt();
int n1=0;
int n2=0;
ArrayList<ArrayList<Integer>> nodeConn = new ArrayList<ArrayList<Integer>>();
int[] nodeLen = new int[n];
for (int i=0; i<n; i++){
nodeConn.add(new ArrayList<Integer>());
}
for (int i=0; i<m; i++){
n1 = in.nextInt()-1;
n2 = in.nextInt()-1;
nodeConn.get(n1).add(n2);
nodeLen[n1]++;
nodeConn.get(n2).add(n1);
nodeLen[n2]++;
}
int min=0;
for (int i=0; i<n; i++){
//System.out.print((i+1) + " "+ nodeLen[i]+" ");
if(nodeLen[i]<nodeLen[min])
min=i;
//System.out.println(min+1);
}
ArrayList<Integer> temp = new ArrayList<Integer>();
int currNode = 0;
int newMin = 0;
int count = 0;
int[] path = new int[n];
while(nodeLen[min]!=0&&count<=n){
path[count]=min+1;
temp = nodeConn.get(min);
newMin=temp.get(0);
for(int i=0; i<temp.size(); i++){
currNode = temp.get(i);
nodeConn.get(currNode).remove((Integer)min);
nodeLen[currNode]--;
if(nodeLen[currNode]<nodeLen[newMin])
newMin=currNode;
}
min = newMin;
count++;
}
if(count!=n){
path[count]=min+1;
count++;
}
System.out.println(count);
for (int i=0; i<count; i++){
System.out.print(path[i] + " ");
}
}
}
In C :
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int max=0,max1=0,index,count;
void DFS(int* V,int i,int n,int* parent,int* g[],int q)
{
//printf(" %s ",node1[i].ch);
V[i] = 1;
count++;
///node* temp;
//temp = node1[i].next;
int j;
for(int j=0;j<n;j++)
{
if(g[i][j]==1&&V[j]==-1){
parent[j] = i;
DFS(V,j,n,parent,g,q+1);
}
}
if(q>max)
{
max = q;
index = i;
}
}
// function for traversal of every non visited nodes
void con_COMP(int size,int* g[])
{
int visited[size],parent[size];
int i;
for(i=0;i<size;i++)
{
visited[i] = -1;
parent[i] = -1;
}
int index1;
for(i = 0;i<size;i++)
{
if(visited[i] == -1)
{
count = 0;
//printf("%d.",count);
DFS(visited,i,size,parent,g,1);
if(count>max1)
{
max1 = count;
index1 = i;
}
// printf("\n");
}
}
printf("%d\n",max);
for(int j = index;j!=index1;j=parent[j])
printf("%d ",j+1);
printf("%d ",index1+1);
}
int main() {
int n,m;
scanf("%d %d",&n,&m);
int** graph = (int**)malloc(sizeof(int*)*(n));
for(int i = 0;i < n;i++)
{
graph[i] = (int*)malloc(sizeof(int)*(n));
}
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
graph[i][j] = 0;
}
for(int i=0;i<m;i++)
{
int v1,v2;
scanf("%d %d",&v1,&v2);
graph[v1-1][v2-1] = 1;
graph[v2-1][v1-1] = 1;
}
con_COMP(n,graph);
return 0;
}
In Python3 :
n,m=map(int,input().split())
roads=[]
dic={}
for i in range(m):
[x,y]=list(map(int,input().split()))
roads.append([x,y])
if not x in dic:dic[x]={y}
if x in dic:dic[x].add(y)
if not y in dic:dic[y]={x}
if y in dic:dic[y].add(x)
count=[0]*(n+1)
for x in dic:
count[x]=len(dic[x])
_,start=min([[count[z],z] for z in dic],key=lambda a:a[0])
path=[start]
x=start
for i in range(1,n):
if len(dic[x])==0:break
_,y=min([[count[z],z] for z in dic[x]],key=lambda a:a[0])
if x in dic[y]: dic[y].remove(x)
count[y]-=1
path.append(y)
x=y
print(len(path))
print(*path, sep=" ")
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