题意:
有n个点,m条边,问能不能找到一条路把所有的边都经过一次。
求解思路:
因为是要遍历所有的边,所以这个问题就可以规约到求一个图的欧拉路径上。
但是这道题的数据是比较严格的,我们不能写递归,因此就要用非递归写欧拉回路,参考博客如下:
https://blog.csdn.net/kk303/article/details/10552441
为了压缩空间使用了向前星,参考博客如下:
https://blog.csdn.net/cryssdut/article/details/24300971
这道题的WA在点于是否判断度数为偶数,如果没有判断会报错。
#include<cstdio>
#include<string.h>
using namespace std;
const int maxn = 4000000+5;
int n, m, tot, num;
int visited[maxn], degree[maxn], head[maxn], path[maxn], s[maxn];
struct Node {
int to;
int next;
}g[maxn];
void addEgde(int u, int v) {
g[tot].to = v; g[tot].next = head[u];
head[u] = tot++;
g[tot].to = u; g[tot].next = head[v];
head[v] = tot++;
}
void Elur(int p) {
int t = 1;
s[t] = p;
while (t) {
int flag = 1;
int v = s[t--];
for (int i = head[v]; ~i; i = g[i].next) {
if (!visited[i]) {
visited[i] = 1; visited[i ^ 1] = 1;
s[++t] = v; s[++t] = g[i].to;
flag = 0;
break;
}
else head[v] = g[i].next;
}
if (flag) path[num++] = v;
}
}
int sum = 0;
int ss = 0;
bool judge() {
for (int i = 0; i < n; i++) {
if (degree[i] & 1) {
sum++; ss = i;
}
}
return sum == 0 || sum == 2;
}
int main() {
memset(head, -1, sizeof(head));
scanf("%d%d", &n, &m);
int a, b;
for (int i = 0; i < m; i++) {
scanf("%d%d", &a, &b);
addEgde(a, b);
degree[a] ++;
degree[b] ++;
}
if (judge()) {
printf("Yes
");
Elur(ss);
for (int i = 0; i < num; i++) {
printf("%d ", path[i]);
}
}
else printf("No
");
return 0;
}