POJ 1041 John's trip 欧拉回路（输出路径）

说出无向图的欧拉回路的路径

使用Fleury算法

（1）其实就是dfs，标记边，逆序输出

（2）保证字典序最小，则每次拓展边时，优先选择边值较小的即可。

（3）因为是逆序输出，所以保存到stack中，在弹出。

下面是2种边的储存法的解法：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <cstdlib>
#include <fstream>
#include <vector>
#include <set>
#include <map>
#include <stack>
#include <algorithm>

using namespace std;

typedef long long LL;
const int INF = 0x3f3f3f3f;
const int maxn = 2010;

struct node{
    int z, x, y;
    node(){}
    node(int z, int x, int y) : z(z), x(x), y(y) {}
    bool operator<(const node &a) const
    {
        return z < a.z;
    }
};
vector<node>adj[2010];
int du[55];
stack<int>S;
int vis[2010];
int en, vn, st;

void dfs(int u)
{
    for (int r = 0; r < adj[u].size(); r++)
    {
        if (!vis[adj[u][r].z])
        {
            int v = adj[u][r].y;
            vis[adj[u][r].z] = 1;
            dfs(v);
            S.push(adj[u][r].z);
        }
    }
}

int dfs2(int u)
{
    vis[u] = 1;
    int ret = 1;
    for (int r = 0; r < adj[u].size(); r++)
    {
        int v = adj[u][r].y;
        if (!vis[v])
            ret += dfs2(v);
    }
    return ret;
}

void solve()
{
    for (int i = 1; i <= vn; i++)
        sort(adj[i].begin(), adj[i].end());
    memset(vis, 0, sizeof(vis));
    int tot = dfs2(st);
    for (int i = 1; i <= vn; i++)
        if (du[i] % 2) tot = 0;

    while (!S.empty()) S.pop();
    memset(vis, 0, sizeof(vis));
    dfs(st);

    if (tot == vn && S.size() == en)
    {
        int p = 0;
        while (!S.empty())
        {
            if (!p) p = 1;
            else printf(" ");
            printf("%d", S.top());
            S.pop();
        }
        puts("");
    }
    else
    {
        puts("Round trip does not exist.");
    }
}
int main()
{
    int x, y, z;
    while (scanf("%d%d",&x,&y) && (x || y))
    {
        scanf("%d", &z);
        memset(du, 0, sizeof(du));
        for (int i = 0; i < 2010; i++)
            adj[i].clear();
        st = min(x, y);
        vn = max(x, y);
        en = z;
        adj[x].push_back(node(z, x, y));
        adj[y].push_back(node(z, y, x));
        du[x]++;
        du[y]++;

        while (scanf("%d%d", &x, &y) && (x || y))
        {
            scanf("%d", &z);
            vn = max(x, vn);
            vn = max(y, vn);
            en = max(en, z);
            adj[x].push_back(node(z, x, y));
            adj[y].push_back(node(z, y, x));
            du[x]++;
            du[y]++;
        }
        solve();
    }
    return 0;
}

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <cstdlib>
#include <fstream>
#include <vector>
#include <set>
#include <map>
#include <stack>

using namespace std;

typedef long long LL;
const int INF = 0x3f3f3f3f;
const int maxn = 2010;

int du[55];
stack<int>S;
int adj[2010][2];
int vis[2010];
int en, vn, st;
void dfs(int u)
{
    for (int i = 1; i <= en; i++)
    {
        if (!vis[i])
        {
            int v;
            if (adj[i][0] == u) v = adj[i][1];
            else if (adj[i][1] == u) v = adj[i][0];
            else continue;
            vis[i] = 1;
            dfs(v);
            S.push(i);
        }
    }
}

int dfs2(int u)
{
    vis[u] = 1;
    int ret = 1;
    for (int i = 1; i <= en; i++)
    {
        int v;
        if (adj[i][0] == u) v = adj[i][1];
        else if (adj[i][1] == u) v = adj[i][0];
        else continue;
        if (!vis[v])
            ret += dfs2(v);
    }
    return ret;
}

void solve()
{
//    cout << vn << ' ' << en << ' ' << st << endl;
    memset(vis, 0, sizeof(vis));
    int tot = dfs2(st);
    for (int i = 1; i <= vn; i++)
        if (du[i] % 2) tot = 0;
//    cout << "^^^^^^^^^^^66" << endl;
//    cout << tot << endl;

    while (!S.empty()) S.pop();
    memset(vis, 0, sizeof(vis));
    dfs(st);
//    cout << "^^^^^^^^^^^66" << endl;
//    cout << S.size() << endl;

    if (tot == vn && S.size() == en)
    {
        int p = 0;
        while (!S.empty())
        {
            if (!p) p = 1;
            else printf(" ");
            printf("%d", S.top());
            S.pop();
        }
        puts("");
    }
    else
    {
        puts("Round trip does not exist.");
    }
}
int main()
{
    int x, y, z;
    while (scanf("%d%d",&x,&y) && (x || y))
    {
        scanf("%d", &z);
        memset(du, 0, sizeof(du));
        st = min(x, y);
        vn = max(x, y);
        en = z;
        adj[z][0] = x; adj[z][1] = y;
        du[x]++;
        du[y]++;

        while (scanf("%d%d", &x, &y) && (x || y))
        {
            scanf("%d", &z);
            vn = max(x, vn);
            vn = max(y, vn);
            en = max(en, z);
            adj[z][0] = x; adj[z][1] = y;
            du[x]++;
            du[y]++;
        }
        solve();
    }

    return 0;
}