The Necklace

题目链接

• 题意：
  给很多由两个颜色组成的珠子，看是否能连成一串，相邻珠子相邻部分颜色相同
• 分析：
  裸地欧拉路判断+路径输出

const int MAXN = 1100;const int MAXV = 60;const int MAXE = 2020;struct Edge{    int from, to;};struct UnDirect_Euler{    int n, m;    int deg[MAXV];    bool vis[MAXE];    vector<int> G[MAXV];    vector<Edge> edges;    stack<int> sk;    vector<int> ans;    void init(int n)    {        this->n = n;        edges.clear();        REP(i, n)        {            deg[i] = 0;            G[i].clear();        }    }    void addEdge(int a, int b)    {        edges.push_back((Edge) { a, b });        edges.push_back((Edge) { b, a });        m = edges.size();        G[a].push_back(m - 2);        G[b].push_back(m - 1);        deg[a]++;        deg[b]++;    }    void dfs(int u, int ind)    {        REP(i, G[u].size())        {            int x = G[u][i];            Edge& e = edges[x];            if (!vis[x])            {                vis[x] = vis[x ^ 1] = true;                dfs(e.to, x);            }        }        if (ind >= 0)            sk.push(ind);    }    //返回1:欧拉回路  返回2：欧拉路经    int solve(int s)    {        ans.clear();        int cnt = 0;        REP(i, n)        {            if (deg[i] == 1)            {                if (++cnt > 2) return 0;                s = i;            }            else if (deg[i] % 2 != 0) return 0;        }        while (!sk.empty()) sk.pop();        REP(i, m) vis[i] = false;        dfs(s, -1);        REP(i, m) if (!vis[i]) return 0;        while (!sk.empty())        {            ans.push_back(sk.top());            sk.pop();        }        return cnt != 0 ? 1 : 2;    }} graph;int a[MAXN], b[MAXN];int main(){    int T, n;    RI(T);    FE(kase, 1, T)    {        graph.init(60);        RI(n);        REP(i, n)        {            RII(a[i], b[i]);            graph.addEdge(a[i], b[i]);        }        printf("Case #%d\n", kase);        if (graph.solve(a[0]) == 2)        {            vector<int>& ans = graph.ans;            REP(i, ans.size())            {                Edge& e = graph.edges[ans[i]];                int u = e.from, v = e.to;                printf("%d %d\n", u, v);            }        }        else puts("some beads may be lost");        puts("");    }    return 0;}