codeforces 557D Vitaly and Cycle

新技能get

二分图染色居然可以用来判断奇偶环。。

如果能构成二分图，说明没有奇环

 

对于任意二分图，其包含的环一定全部是偶环！（充要可证）

可以证明，含有奇数条边的环一定有两个在相同集合内的点有边相连！

也就是说——二分图的bfs子树一定不含奇环！

 

下面讨论摘自http://www.cnblogs.com/sagitta/p/4612214.html

 

首先讨论添加边条数为3——即原原图边数m为0时

ns=n*(n-1)*(n-2)/6

再讨论添加边条数为2——即原图中所有边都没有公共端点/所有点度数<=1 时

ans=m*(n-2)

再讨论添加边数为0——即原图中存在奇环时

ans=1

最后讨论添加边数为1——即原图中只有树以及偶环

ans=Σ[ (white[i]-1)*white[i]/2+(black[i]-1)*black[i]/2 ]

 

看的matrix大牛代码

dfs中直接求ans

 1 /*Author :usedrose  */
 2 /*Created Time :2015/7/19 17:01:15*/
 3 /*File Name :2.cpp*/
 4 #include <cstdio>
 5 #include <iostream>
 6 #include <algorithm>
 7 #include <sstream>
 8 #include <cstdlib>
 9 #include <cstring>
10 #include <climits>
11 #include <vector>
12 #include <string>
13 #include <ctime>
14 #include <cmath>
15 #include <deque>
16 #include <queue>
17 #include <stack>
18 #include <set>
19 #include <map>
20 #define INF 0x3f3f3f3f
21 #define eps 1e-8
22 #define pi acos(-1.0)
23 #define MAXN 100110
24 #define OK cout << "ok" << endl;
25 #define o(a) cout << #a << " = " << a << endl
26 #define o1(a,b) cout << #a << " = " << a << "  " << #b << " = " << b << endl
27 using namespace std;
28 typedef long long LL;
29 
30 int n, m;
31 bool marked[MAXN], ok;
32 int  sz[2];
33 int col[MAXN];
34 vector<int > G[MAXN];
35 LL ans;
36 
37 void dfs(int v, int c)
38 {
39     marked[v] = true;
40     ans += sz[c-1];
41     sz[c-1]++;
42     col[v] = c;
43     for (int i = 0; i < G[v].size(); i++)
44     {
45         int u = G[v][i];
46         if (!marked[u])
47         {
48             col[u] = 3 - c;
49             dfs(u, 3 - c);
50         }
51         else if (col[u] == col[v])
52             ok = true;
53     }
54 }
55 int main()
56 {
57     //freopen("data.in","r",stdin);
58     //freopen("data.out","w",stdout);
59     cin.tie(0);
60     ios::sync_with_stdio(false);
61     cin >> n >> m;
62     int a, b;
63 
64     for (int i = 0;i < m; ++ i) {
65         cin >> a >> b;
66         G[a].push_back(b);
67         G[b].push_back(a);
68     }
69     for (int i = 1; i <= n; ++ i)
70         if (!marked[i]) {
71             sz[0] = sz[1] = 0;
72             dfs(i, 1);
73         }
74     if (ok)
75         cout << 0 << " " << 1 << endl;
76     else if (ans == 0)
77     {
78         if (m == 0)
79         {
80             long long ans = n;
81             ans *= n - 1;
82             ans *= n - 2;
83             ans /= 6;
84             cout << 3 << " " << ans << endl;
85         }
86         else
87             cout << 2 << " " << (long long)m * (n - 2) << endl;
88 
89     }
90     else
91         cout << 1 << " " << ans << endl;
92     
93     return 0;
94 }
View Code

 

 

待做。。 

http://www.hardbird.net/hdu-5215-cycle%E4%BA%8C%E5%88%86%E5%9B%BE%E6%9F%93%E8%89%B2%E5%88%A4%E5%A5%87%E5%81%B6%E7%8E%AF/

posted @ 2015-07-19 17:55  UsedRose  阅读(341)  评论(0)  编辑  收藏  举报