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):

 1 #include <set>
 2 #include <map>
 3 #include <queue>
 4 #include <stack>
 5 #include <cmath>
 6 #include <bitset>
 7 #include <cstdio>
 8 #include <string>
 9 #include <vector>
10 #include <cstdlib>
11 #include <cstring>
12 #include <iostream>
13 #include <algorithm>
14 using namespace std;
15 
16 typedef long long ll;
17 typedef pair<ll, ll> pll;
18 typedef pair<ll, int> pli;
19 typedef pair<int, ll> pil;;
20 typedef pair<int, int> pii;
21 typedef unsigned long long ull;
22 
23 #define lson i<<1
24 #define rson i<<1|1
25 #define bug printf("*********\n");
26 #define FIN freopen("D://code//in.txt", "r", stdin);
27 #define debug(x) cout<<"["<<x<<"]" <<endl;
28 #define IO ios::sync_with_stdio(false),cin.tie(0);
29 
30 const double eps = 1e-8;
31 const int mod = 10007;
32 const int maxn = 5e4 + 7;
33 const double pi = acos(-1);
34 const int inf = 0x3f3f3f3f;
35 const ll INF = 0x3f3f3f3f3f3f3f;
36 
37 int n, m, u, v, tot, top, t;
38 int head[maxn], stc[maxn*10], ans[maxn*10], vis[maxn*10];
39 
40 struct edge {
41     int v, next;
42 }ed[maxn*10];
43 
44 void addedge(int u, int v) {
45     ed[tot].v = v;
46     ed[tot].next = head[u];
47     head[u] = tot++;
48     ed[tot].v = u;
49     ed[tot].next = head[v];
50     head[v] = tot++;
51 }
52 
53 void eulergraph() {
54     stc[++top] = 1;
55     while(top > 0) {
56         int x = stc[top], i = head[x];
57         while(i != -1 && vis[i]) i = ed[i].next;
58         if(i != -1) {
59             stc[++top] = ed[i].v;
60             head[x] = ed[i].next;
61             vis[i] = 1;
62         } else {
63             top--;
64             ans[++t] = x;
65         }
66     }
67 }
68 
69 int main() {
70     //FIN;
71     scanf("%d%d", &n, &m);
72     tot = top = t = 0;
73     memset(stc, 0, sizeof(stc));
74     memset(ans, 0, sizeof(ans));
75     memset(vis, 0, sizeof(vis));
76     memset(head, -1, sizeof(head));
77     for(int i = 1; i <= m; i++) {
78         scanf("%d%d", &u, &v);
79         addedge(u, v);
80     }
81     eulergraph();
82     for(int i = t; i; i--) printf("%d\n", ans[i]);
83 }

 

 1 #include <set>
 2 #include <map>
 3 #include <queue>
 4 #include <stack>
 5 #include <cmath>
 6 #include <bitset>
 7 #include <cstdio>
 8 #include <string>
 9 #include <vector>
10 #include <cstdlib>
11 #include <cstring>
12 #include <iostream>
13 #include <algorithm>
14 using namespace std;
15 
16 typedef long long ll;
17 typedef pair<ll, ll> pll;
18 typedef pair<ll, int> pli;
19 typedef pair<int, ll> pil;;
20 typedef pair<int, int> pii;
21 typedef unsigned long long ull;
22 
23 #define lson i<<1
24 #define rson i<<1|1
25 #define bug printf("*********\n");
26 #define FIN freopen("D://code//in.txt", "r", stdin);
27 #define debug(x) cout<<"["<<x<<"]" <<endl;
28 #define IO ios::sync_with_stdio(false),cin.tie(0);
29 
30 const double eps = 1e-8;
31 const int mod = 10007;
32 const int maxn = 1e4 + 7;
33 const double pi = acos(-1);
34 const int inf = 0x3f3f3f3f;
35 const ll INF = 0x3f3f3f3f3f3f3f;
36 
37 int n, m, u, v, tot, top, t;
38 int head[maxn], vis[maxn*10];
39 
40 struct edge {
41     int v, next;
42 }ed[maxn*10];
43 
44 void addedge(int u, int v) {
45     ed[tot].v = v;
46     ed[tot].next = head[u];
47     head[u] = tot++;
48     ed[tot].v = u;
49     ed[tot].next = head[v];
50     head[v] = tot++;
51 }
52 
53 void eulergraph(int x) {
54     for(int i = head[x]; ~i; i = ed[i].next) {
55         if(!vis[i]) {
56             vis[i] = 1;
57             eulergraph(ed[i].v);
58         }
59     }
60     printf("%d\n", x);
61 }
62 
63 int main() {
64     //FIN;
65     scanf("%d%d", &n, &m);
66     tot = top = t = 0;
67     memset(vis, 0, sizeof(vis));
68     memset(head, -1, sizeof(head));
69     for(int i = 1; i <= m; i++) {
70         scanf("%d%d", &u, &v);
71         addedge(u, v);
72     }
73     eulergraph(1);
74     return 0;
75 }

 

posted @ 2018-07-31 14:37  Dillonh  阅读(271)  评论(0)  编辑  收藏  举报