Codeforces 29C Mail Stamps 【离散化 + DFS】

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题目链接：Codeforces 29C Mail Stamps

题意：给你n条无向边，保证只存在两条路径可以遍历完所有的点。让你输出任意一条路径。

AC代码：

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <algorithm>#include <vector>#include <queue>#include <map>#include <stack>#define PI acos(-1.0)#define CLR(a, b) memset(a, (b), sizeof(a))#define fi first#define se second#define ll o<<1#define rr o<<1|1using namespace std;typedef long long LL;typedef pair<int, int> pii;const int MAXN = 2*1e5+100;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;void getmax(int &a, int b) {a = max(a, b); }void getmin(int &a, int b) {a = min(a, b); }void add(LL &x, LL y) { x += y; x %= MOD; }int u[MAXN], v[MAXN];map<int, int> fp;vector<int> G[MAXN];bool vis[MAXN];int rec[MAXN];void DFS(int u){    for(int i = 0; i < G[u].size(); i++)    {        int v = G[u][i];        if(vis[v]) continue;        cout << " " << rec[v];        vis[v] = true;        DFS(v);    }}int main(){    int n; cin >> n; int top = 0;    fp.clear();    for(int i = 1; i <= n; i++)    {        scanf("%d%d", &u[i], &v[i]);        rec[top++] = u[i];        rec[top++] = v[i];        fp[u[i]]++; fp[v[i]]++;    }    sort(rec, rec+top);    int T = unique(rec, rec+top) - rec;    //cout << T << endl;    for(int i = 0; i < T; i++) {        G[i].clear(); vis[i] = false;    }    for(int i = 1; i <= n; i++) {        int sx = lower_bound(rec, rec+T, u[i]) - rec;        int ex = lower_bound(rec, rec+T, v[i]) - rec;        G[sx].push_back(ex);        G[ex].push_back(sx);    }    int s;    for(int i = 0; i < T; i++) {        if(fp[rec[i]] == 1) {            s = i;            break;        }    }    cout << rec[s]; vis[s] = true; DFS(s);    return 0;}

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