hihoCoder-1181-Oralu Shi (Fleury algorithm to find Euler path)

#1181: Oralu Shi time limit:10000msSingle Point time limit:1000msMemory Limit:256MB

Describe

In the last back small hi and small ho control the protagonist gathered scattered on the wooden bridge props, these props are actually a piece of dominoes.

The protagonist continued to move forward, a stone bridge appeared in front of the stone bridge at the end of a fire wall, it seems unable to pass.

Little Hi noticed that there was a small piece of paper at the bridge, so the controlling protagonist picked up the paper, and saw it read:

The flame wall can be closed by placing the M-block dominoes in the concave of the stone bridge. Remember that dominoes need the same number to connect. --by unknown adventurer.

Small hi and Small ho opened the main character of the props bar, found that the protagonist happens to have M fast Domino.

Small ho: That means to put all the dominoes in the groove to close the flame wall, what does the number mean?

Little hi: You see, each piece of dominoes has a number at each end, presumably only when the number is the same, can be placed together, such as:

Little ho: So, let's see if we can connect all the dominoes together.

Hint: Fleury algorithm to find Euler path

Input

Line 1th: 2 positive integers, n,m. Indicates the maximum number and number of dominoes that appear on the dominoes, respectively. 1≤n≤1,000,1≤m≤5,000

2nd.. M+1 lines: 2 integers per line, u,v. Line I+1 represents the number (U,V) at both ends of the block I, 1≤u,v≤n

Output

Line 1th: m+1 numbers, indicating the number after the end of a domino

For example, the status of the Domino connection is (1,5) (5,3) (3,2) (2,4) (4,3), then output "1 5 3 2 4 3"

You can output any set of legitimate solutions.

Sample input

5 53 53 24 23 45 1

Sample output

1 5 3 4 2 3

Fleury algorithm pseudo-code:

DFS (U): while (U exists not deleted Edge E (u,v)) Delete Edge E (u,v) DFS (v) endpathsize←pathsize + 1path[pathsize]←u

Here to notice how to delete the edge, you can use the vector's erase, you can also use a marker to determine whether the edge has been deleted, such as the edge to delete the value of 1.

Deletion of vectors:

#include <vector> #include <iostream>using namespace std; int main (int argc, char** argv) {    std::vector<int> vec;    for (int i=0;i<100;i++)    {         vec.push_back (i);    }     printf ("10:%d\n", vec[10]);    printf ("size:%d\n", Vec.size ());    printf ("**********************************\n");    Std::vector<int>::iterator it = Vec.begin () +10;    Vec.erase (it);     printf ("10:%d\n", vec[10]);    printf ("size:%d\n", Vec.size ());    return 0;}   Output//10:10//size:100//**********************************//10:11//size:99

AC Code (ERASE):

#include <map> #include <set> #include <cmath> #include <deque> #include <queue> #include <stack> #include <cstdio> #include <cctype> #include <string> #include <vector> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> #define LL long long# Define INF 0x7fffffffusing namespace Std;int n, m;const int maxn = 1005;vector<int> g[maxn];int path[5005];int paths ize;void dfs (int u) {int D;while ((d = g[u].size ()) > 0) {int v = g[u][0];//cout << v << "" << D < < Endl;  G[u].erase (G[u].begin ()); int L = G[v].size (); for (int i = 0; I < L + +) {//Find the connection of V to u with u, because both sides are mutual, to remove if (g[v][i] = = u) {g[v].erase (G[v].begin () + i); break;}} DFS (v);} Path[pathsize + +] = u;} int main () {while (scanf ("%d%d", &n, &m)! = EOF) {for (int i = 0; i < m; i + +) {int u, v;scanf ("%d%d", &u, &AMP;V); G[u].push_back (v); G[v].push_back (u);} Pathsize = 0;int start = 1;while (! G[start].sIze ()) Start ++;//Find the first point with a connecting edge, but here the data is not so, later notice DFS (start);//Here Start write 1 can also AC for (int i = 0; i < pathsize-1; i + +) {Prin TF ("%d", Path[i]);} printf ("%d\n", Path[pathsize-1]);} return 0;}

AC Code (Mark):

#include <map> #include <set> #include <cmath> #include <deque> #include <queue> #include <stack> #include <cstdio> #include <cctype> #include <string> #include <vector> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> #define LL long long# Define INF 0x7fffffffusing namespace Std;int n, m;const int maxn = 1005;vector<int> g[maxn];int path[5005];int paths  ize;void dfs (int u) {int d = g[u].size (); for (int i = 0; i < D; i + +) {int v = g[u][i];if (v! =-1) {G[u][i] = -1;int L = G[v].size (); for (int j = 0; J < L; j + +) {//Find the connecting edge of V connected to u, because both sides are mutual, to remove if (g[v][j] = = u) {g[v][j] = -1;break;}} DFS (v);}} Path[pathsize + +] = u;} int main () {while (scanf ("%d%d", &n, &m)! = EOF) {for (int i = 0; i < m; i + +) {int u, v;scanf ("%d%d", &u, &AMP;V); G[u].push_back (v); G[v].push_back (u);} int start = 1;while (! G[start].size ()) Start + +;p athsize = 0;dfs (start); for (int i = 0; i < pathsize-1; i + +) {printf ("%d", Path[i]);} printf ("%d\n", Path[pathsize-1]);} return 0;}

Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced.

hihoCoder-1181-Oralu Shi (Fleury algorithm to find Euler path)