例题6-16 单词（Play On Words, UVa 10129）

裸的有向图判欧拉通路问题。。
所有点入度 = 出度则存在欧拉回路，度数不等点恰好为2，其中一点且入度 - 出度 = 1，则为欧拉通路。

#include <iostream>#include <string>#include <vector>#include <stack>#include <queue>#include <deque>#include <set>#include <map>#include <algorithm>#include <functional>#include <utility>#include <cstring>#include <cstdio>#include <cstdlib>#include <ctime>#include <cmath>#include <cctype>#define CLEAR(a, b) memset(a, b, sizeof(a))#define IN() freopen("in.txt", "r", stdin)#define OUT() freopen("out.txt", "w", stdout)#define LL long long#define maxn 1005#define maxm 100005#define mod  10007#define INF 1000000007#define EPS 1e-7#define PI 3.1415926535898#define N 4294967296using namespace std;//-------------------------CHC------------------------------//typedef pair<int, int> pii;char s[maxn];int in[26], out[26];bool have_node[26];vector<int> edges;vector<int> G[26];bool check(vector<pii> &v) {return (v[0].first - v[0].second) * (v[1].first - v[1].second) == -1;}void add(int u, int v) {edges.push_back(v);G[u].push_back(edges.size() - 1);}bool vis[26];void dfs(int u) {if (vis[u]) return;vis[u] = true;for (int i = 0; i < G[u].size(); ++i)dfs(edges[G[u][i]]);}int main() {//IN(); OUT();int T;scanf("%d", &T);while (T--) {for (int i = 0; i < 26; ++i) G[i].clear();CLEAR(in, 0);CLEAR(out, 0);CLEAR(have_node, 0);CLEAR(vis, 0);int n;scanf("%d", &n);for (int i = 0; i < n; ++i) {scanf("%s", s);int u = s[strlen(s) - 1] - 'a';int v = s[0] - 'a';have_node[u] = have_node[v] = true;in[u]++, out[v]++;add(u, v);add(v, u);}vector<pii> dif;for (int i = 0; i < 26; ++i) {if (in[i] != out[i]) dif.push_back(pii(in[i], out[i]));}dfs(s[0] - 'a');bool ok = true;for(int i = 0; i < 26; ++i) if (have_node[i] && !vis[i]) ok = false;if (!ok) { puts("The door cannot be opened."); continue; }if (dif.size() == 0) puts("Ordering is possible.");else if (dif.size() == 2 && check(dif)) puts("Ordering is possible.");else puts("The door cannot be opened.");}return 0;}

用并查集判连通，无需建图，只需记录连通分量的个数，初始为点的个数，随着unite而减少。
入度出度用同一个数组表示，正负来区分。
在度数不为零数为2时，由握手定理的入度等于出度，故只需要判断一个点的度数为1或-1。

#include <iostream>#include <string>#include <vector>#include <stack>#include <queue>#include <deque>#include <set>#include <map>#include <algorithm>#include <functional>#include <utility>#include <cstring>#include <cstdio>#include <cstdlib>#include <ctime>#include <cmath>#include <cctype>#define CLEAR(a, b) memset(a, b, sizeof(a))#define IN() freopen("in.txt", "r", stdin)#define OUT() freopen("out.txt", "w", stdout)#define LL long long#define maxn 1005#define maxm 100005#define mod  10007#define INF 1000000007#define EPS 1e-7#define PI 3.1415926535898#define N 26using namespace std;//-------------------------CHC------------------------------//char s[maxn];int degree[maxn];bool have_node[N];int f[N], cnt;void init() {cnt = N;CLEAR(have_node, 0);CLEAR(degree, 0);for (int i = 0; i < N; ++i) f[i] = i;}int find(int x) { return x == f[x] ? x : f[x] = find(f[x]); }void unite(int a, int b) {int fa = find(a), fb = find(b);if (fa == fb) return;f[fa] = fb;cnt--;}bool check(vector<int> &dif) {bool ret = false;if (dif.empty()) ret = true;else if (dif.size() == 2) {if (dif[0] == 1 || dif[0] == -1) ret = true;}return ret;}int main() {int T;scanf("%d", &T);while (T--) {init();int n;scanf("%d", &n);for (int i = 0; i < n; ++i) {scanf("%s", s);int u = s[strlen(s) - 1] - 'a';int v = s[0] - 'a';have_node[u] = have_node[v] = true;degree[u]++, degree[v]--;unite(u, v);}vector<int> dif;for (int i = 0; i < N; ++i) {if (!have_node[i]) cnt--;else if (degree[i]) dif.push_back(degree[i]);}bool ok = false;ok = (cnt == 1 && check(dif));puts(ok ? "Ordering is possible." : "The door cannot be opened.");}return 0;}