HDU 6184 Counting Stars

Posted 2021-01-15 mjtcn

Counting Stars

http://acm.hdu.edu.cn/showproblem.php?pid=6184

题意：求这样图形的个数。

分析：

　　三元环计数。

　　两个三元环可以组成一个那样的图形。于是直接枚举一条边，然后求这条边所能构成的三元环。

　　三元环的求法和更优的做法

　　更优的做法

代码：

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<iostream>
 6 #include<cctype>
 7 #include<set>
 8 #include<vector>
 9 #include<queue>
10 #include<map>
11 #define fi(s) freopen(s,"r",stdin);
12 #define fo(s) freopen(s,"w",stdout);
13 using namespace std;
14 typedef long long LL;
15 
16 inline int read() {
17     int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch==‘-‘)f=-1;
18     for(;isdigit(ch);ch=getchar())x=x*10+ch-‘0‘;return x*f;
19 }
20 
21 const int N = 100005;
22 
23 vector<int>e[N];
24 set<LL> s;
25 int A[N], B[N], deg[N], Link[N];
26 bool vis[N];
27 
28 int main() {
29     int n, m;
30     while (~scanf("%d%d",&n, &m)) {
31         int Size = sqrt(m);
32         s.clear();
33         for (int i = 1; i <= n; ++i) Link[i] = deg[i] = 0, vis[i] = false, e[i].clear();
34         for (int i = 1; i <= m; ++i) {
35             int u = read(), v = read();
36             deg[u] ++, deg[v] ++;
37             e[u].push_back(v), e[v].push_back(u);
38             s.insert(1ll * u * n + v), s.insert(1ll * v * n + u);
39         }
40         LL Ans = 0;
41         for (int u = 1; u <= n; ++u) {
42             vis[u] = true; // 如果没有vis，Ans需要/2 
43             for (int i = 0; i < e[u].size(); ++i) Link[e[u][i]] = u;
44             for (int i = 0; i < e[u].size(); ++i) {
45                 int v = e[u][i], cnt = 0;
46                 if (vis[v]) continue;
47                 if (deg[v] <= Size) {
48                     for (int j = 0; j < e[v].size(); ++j) 
49                         if (Link[e[v][j]] == u) cnt ++;
50                 }
51                 else {
52                     for (int j = 0; j < e[u].size(); ++j) 
53                         if (s.find(1ll * e[u][j] * n + v) != s.end()) cnt ++;
54                 }
55                 Ans += (1ll * cnt * (cnt - 1) / 2);
56             }
57         }
58         printf("%lld
",Ans);
59     }
60     return 0;
61 }