sgu 156 特殊情况下的哈密顿路

仔细读题和推敲可以发现，原图是由许多完全图由许多度为2的节点连接起来的。

如果将这些完全图缩成一个个的点，此时图完全是由许多2度点组成的，那么这时的哈密顿路求解就可以转化成欧拉路的求解。

需要仔细考虑的是将新图生成的欧拉路还原成原图的哈密顿路的过程

// sgu156
const int Debug = 1;
/*
 * SOUR:
 * ALGO:
 * DATE:2011 6月06 13时20分47秒
 * COMM:
 * */
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cassert>
#include<list>
#include<cmath>
#include<queue>
#include<vector>
#include<map>
#include<stack>
#include<set>
using namespace std;
#define pb(x) push_back(x)
#define fi first
#define se second
#define rab(i,a,b) for(int i(a), _n(b); i <= (_n); i++)
#define rep(i,n) rab(i,0,(n)-1)

typedef std::vector < int >vi;
typedef std::pair < int, int > pii;
typedef unsigned int uint;
typedef long long LL;

template <class T> void ckmin(T &a,T b) { if (a > b) { a = b; } }
template <class T> void ckmax(T &a,T b) { if (a < b) { a = b; } }
template <class T> void pr(T &a) { cout << a << ' '; }
template <class T> void print(T &a) { a.__str__(); }
template <class T> int size(const vector<T> &v) { return (int)v.size(); }
#define fpr(...) \
    fprintf(stderr, "%s(%d)-%s: ",__FILE__,__LINE__,__FUNCTION__); \
    fprintf(stderr, __VA_ARGS__);fprintf(stderr, "\n");
int countbit(int n) { return n == 0 ? 0 : 1 + countbit(n & (n - 1)); } 
const int maxint = 0x7fffffff;
const long long max64 = 0x7fffffffffffffffll;
/*Every problem has a simple, fast and wrong solution.*/
/*std::ios::sync_with_stdio(false);*/

const int N = 10001;
const int M = 100010;

int n, m, vn;
int lab[N], deg[N];
int path[M + 100];

vector<int>  adj[N];
vector<int> adj2[N];
uint beg[N];
int bal[N];
map<int, int> vis[N];


int len;
bool euler(int u)
{
    for (uint i = 0; i < adj2[u].size(); i++) {
        const int &v = adj2[u][i];
        if (!vis[u][v]) {
            vis[u][v] = true;
            vis[v][u] = true;

            euler(v);
        }
    }
    path[len++] = u;
}

int touched[N];
void dfs(int u)
{
    touched[u] = true;
    for (uint i = 0; i < adj[u].size(); i++) {
        const int &v = adj[u][i];
        if (!touched[v]) {
            dfs(v);
        }
    }
}

bool isConnected()
{
    fill(touched, touched + N, 0);
    dfs(1);
    for (int i = 1; i <= n; i++) {
        if (!touched[i]) {
            return false;
        }
    }
    return true;
}

int main()
{
    scanf("%d %d", &n, &m);
    for (int i = 0; i < m; i++) {
        int u, v;
        scanf("%d %d", &u, &v);
        adj[u].push_back(v);
        adj[v].push_back(u);
        deg[u]++;
        deg[v]++;
    }

    for (int i = 1; i <= n; i++) if (lab[i] == 0) {
            lab[i] = ++vn;
            if (deg[i] > 2) {
                for (uint j = 0; j < adj[i].size(); j++) if (deg[adj[i][j]] > 2) {
                        lab[adj[i][j]] = vn;
                    }
            }
        }

    for (int i = 1; i <= n; i++) {
        if (deg[i] == 2) {
            bal[lab[i]] = i;
        } else {
            bal[lab[i]] = -1;
        }
        for (uint j = 0; j < adj[i].size(); j++) if (lab[i] != lab[adj[i][j]]) { //except the self loops
                adj2[lab[i]].push_back(lab[adj[i][j]]);
            }
    }

    bool has_euler_tour = true;
    int edge_num = 0;
    for (int i = 1; i <= vn; i++) {
        edge_num += adj2[i].size();

        if (adj2[i].size() % 2 == 1) {
            has_euler_tour = false;
            break;
        }
    }

    if (!isConnected()) {
        has_euler_tour = false;
    }

    if (!has_euler_tour) {
        puts("-1");
        return 0;
    }

    euler(1);

    edge_num /= 2; 
    assert(len == edge_num + 1);

    for (int i = 0; i < edge_num; i++) {
        const int &u = path[i];
        if (bal[u] < 0) {
            const int ul = bal[path[(i + edge_num - 1) % edge_num]];
            const int ur = bal[path[(i + edge_num + 1) % edge_num]];
            assert(ul > 0 && ur > 0);

            for (uint j = 0; j < 2; j++) {
                for (uint k = 0; k < 2; k++) {
                    if (lab[adj[ul][j]] == u && u == lab[adj[ur][k]]) {
                        printf("%d %d ", adj[ul][j], adj[ur][k]);
                    }
                }
            }
        } else {
            printf("%d ", bal[u]);
        }
    }
    putchar(10);

    return 0;
}