acwing算法提高之图论--欧拉回路和欧拉路径

1 介绍

本专题用来记录欧拉回路和欧拉路径相关的题目。

相关结论：
（1）对于无向图，所有边都是连通的。
（1.1）存在欧拉路径的充要条件：度数为奇数的结点只能是0个或者2个。
（1.2）存在欧拉回路的充要条件：度数为奇数的结点只能是0个。

（2）对于有向图，所有边都是连通的。
（2.1）存在欧拉路径的充要条件1：所有结点的出度均等于其入度。
（2.1）存在欧拉路径的充要条件2：除去两个结点外，其余所有结点的出度等于入度。且除去的那两个结点，其中一个结点的出度比入度多1（起点），另一个结点的入度比出度多1（终点）。
（2.2）存在欧拉回路的充要条件：所有结点的出度均等于其入度。

2 训练

题目1：1123铲雪车

C++代码如下，

#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>using namespace std;int main() {double x1, y1, x2, y2;cin >> x1 >> y1;double sum = 0;while (cin >> x1 >> y1 >> x2 >> y2) {double dx = x1 - x2;double dy = y1 - y2;sum += sqrt(dx * dx + dy * dy) * 2;}int minutes = round(sum / 1000 / 20 * 60);int hours = minutes / 60;minutes %= 60;printf("%d:%02d\n", hours, minutes);return 0;
}

题目2：1184欧拉回路

C++代码如下，

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>using namespace std;const int N = 100010, M = 400010;int type;
int n, m;
int h[N], e[M], ne[M], idx;
bool used[M];
int ans[M], cnt;
int din[N], dout[N];void add(int a, int b) {e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}void dfs(int u) {for (int &i = h[u]; ~i;) {if (used[i]) {i = ne[i];continue;}used[i] = true;if (type == 1) used[i ^ 1] = true;int t;if (type == 1) {t = i / 2 + 1;if (i & 1) t = -t;} else {t = i + 1;}int j = e[i];i = ne[i];dfs(j);ans[++cnt] = t;}
}int main() {scanf("%d", &type);scanf("%d%d", &n, &m);memset(h, -1, sizeof h);for (int i = 0; i < m; ++i) {int a, b;scanf("%d%d", &a, &b);add(a, b);if (type == 1) add(b, a);din[b]++, dout[a]++;}if (type == 1) {for (int i = 1; i <= n; ++i) {if (din[i] + dout[i] & 1) {puts("NO");return 0;}}} else {for (int i = 1; i <= n; ++i) {if (din[i] != dout[i]) {puts("NO");return 0;}}}for (int i = 1; i <= n; ++i) {if (h[i] != -1) {dfs(i);break;}}if (cnt < m) {puts("NO");return 0;}puts("YES");for (int i = cnt; i; --i) printf("%d ", ans[i]);puts("");return 0;
}

题目3：1124骑马修栅栏

C++代码如下，

#include <cstring>
#include <iostream>
#include <algorithm>using namespace std;const int N = 510;int n = 500, m;
int g[N][N];
int ans[1100], cnt;
int d[N];void dfs(int u) {for (int i = 1; i <= n; ++i) {if (g[u][i]) {g[u][i]--, g[i][u]--;dfs(i);}}ans[++cnt] = u;
}int main() {cin >> m;while (m--) {int a, b;cin >> a >> b;g[a][b]++, g[b][a]++;d[a]++, d[b]++;}int start = 1;while (!d[start]) start++;for (int i = 1; i <= n; ++i) {if (d[i] % 2) {start = i;break;}}dfs(start);for (int i = cnt; i; i--) printf("%d\n", ans[i]);return 0;
}

题目4：1185单词游戏

C++代码如下，

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>using namespace std;const int N = 30;int n;
int din[N], dout[N], p[N];
bool st[N];int find(int x) {if (p[x] != x) p[x] = find(p[x]);return p[x];
}int main() {char str[1010];int T;scanf("%d", &T);while (T--) {scanf("%d", &n);memset(din, 0, sizeof din);memset(dout, 0, sizeof dout);memset(st, 0, sizeof st);for (int i = 0; i < 26; ++i) p[i] = i;for (int i = 0; i < n; ++i) {scanf("%s", str);int len = strlen(str);int a = str[0] - 'a', b = str[len - 1] - 'a';st[a] = st[b] = true;dout[a]++, din[b]++;p[find(a)] = find(b);}int start = 0, end = 0;bool success = true;for (int i = 0; i < 26; ++i) {if (din[i] != dout[i]) {if (din[i] == dout[i] + 1) end++;else if (din[i] + 1 == dout[i]) start++;else {success = false;break;}}}if (success && !(!start && !end || start == 1 && end == 1)) success = false;int rep = -1;for (int i = 0; i < 26; ++i) {if (st[i]) {if (rep == -1) rep = find(i);else if (rep != find(i)) {success = false;break;}}}if (success) puts("Ordering is possible.");else puts("The door cannot be opened.");}return 0;
}