Watchcow（POJ2230+双向欧拉回路+打印路径）

题目链接：http://poj.org/problem?id=2230

题目：

题意：给你m条路径，求一条路径使得从1出发最后回到1，并满足每条路径都恰好被沿着正反两个方向经过一次。

思路：由于可以回到起点，并且题目保证有解，所以本题是欧拉回路。输出路径有两种方法，一种是递归实现，一种是用栈处理，不过两种速度差了1s，不知道是不是我递归没处理好~

代码实现如下（第一种为栈输出路径，对应797ms，第二种为递归，对应1782ms):

 #include <set>
 #include <map>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <bitset>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <cstdlib>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 typedef long long ll;
 typedef pair<ll, ll> pll;
 typedef pair<ll, int> pli;
 typedef pair<int, ll> pil;;
 typedef pair<int, int> pii;
 typedef unsigned long long ull;

 #define lson i<<1
 #define rson i<<1|1
 #define bug printf("*********\n");
 #define FIN freopen("D://code//in.txt", "r", stdin);
 #define debug(x) cout<<"["<<x<<"]" <<endl;
 #define IO ios::sync_with_stdio(false),cin.tie(0);

 const double eps = 1e-;
 const int mod = ;
 const int maxn = 5e4 + ;
 const double pi = acos(-);
 const int inf = 0x3f3f3f3f;
 const ll INF = 0x3f3f3f3f3f3f3f;

 int n, m, u, v, tot, top, t;
 int head[maxn], stc[maxn*], ans[maxn*], vis[maxn*];

 struct edge {
     int v, next;
 }ed[maxn*];

 void addedge(int u, int v) {
     ed[tot].v = v;
     ed[tot].next = head[u];
     head[u] = tot++;
     ed[tot].v = u;
     ed[tot].next = head[v];
     head[v] = tot++;
 }

 void eulergraph() {
     stc[++top] = ;
     while(top > ) {
         int x = stc[top], i = head[x];
         while(i != - && vis[i]) i = ed[i].next;
         if(i != -) {
             stc[++top] = ed[i].v;
             head[x] = ed[i].next;
             vis[i] = ;
         } else {
             top--;
             ans[++t] = x;
         }
     }
 }

 int main() {
     //FIN;
     scanf("%d%d", &n, &m);
     tot = top = t = ;
     memset(stc, , sizeof(stc));
     memset(ans, , sizeof(ans));
     memset(vis, , sizeof(vis));
     memset(head, -, sizeof(head));
     for(int i = ; i <= m; i++) {
         scanf("%d%d", &u, &v);
         addedge(u, v);
     }
     eulergraph();
     for(int i = t; i; i--) printf("%d\n", ans[i]);
 }

 #include <set>
 #include <map>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <bitset>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <cstdlib>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 typedef long long ll;
 typedef pair<ll, ll> pll;
 typedef pair<ll, int> pli;
 typedef pair<int, ll> pil;;
 typedef pair<int, int> pii;
 typedef unsigned long long ull;

 #define lson i<<1
 #define rson i<<1|1
 #define bug printf("*********\n");
 #define FIN freopen("D://code//in.txt", "r", stdin);
 #define debug(x) cout<<"["<<x<<"]" <<endl;
 #define IO ios::sync_with_stdio(false),cin.tie(0);

 const double eps = 1e-;
 const int mod = ;
 const int maxn = 1e4 + ;
 const double pi = acos(-);
 const int inf = 0x3f3f3f3f;
 const ll INF = 0x3f3f3f3f3f3f3f;

 int n, m, u, v, tot, top, t;
 int head[maxn], vis[maxn*];

 struct edge {
     int v, next;
 }ed[maxn*];

 void addedge(int u, int v) {
     ed[tot].v = v;
     ed[tot].next = head[u];
     head[u] = tot++;
     ed[tot].v = u;
     ed[tot].next = head[v];
     head[v] = tot++;
 }

 void eulergraph(int x) {
     for(int i = head[x]; ~i; i = ed[i].next) {
         if(!vis[i]) {
             vis[i] = ;
             eulergraph(ed[i].v);
         }
     }
     printf("%d\n", x);
 }

 int main() {
     //FIN;
     scanf("%d%d", &n, &m);
     tot = top = t = ;
     memset(vis, , sizeof(vis));
     memset(head, -, sizeof(head));
     for(int i = ; i <= m; i++) {
         scanf("%d%d", &u, &v);
         addedge(u, v);
     }
     eulergraph();
     return ;
 }