HDU 6184 Counting Stars

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：

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<iostream>
#include<cctype>
#include<set>
#include<vector>
#include<queue>
#include<map>
#define fi(s) freopen(s,"r",stdin);
#define fo(s) freopen(s,"w",stdout);
using namespace std;
typedef long long LL;

inline int read() {
    int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
    for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
}

const int N = 100005;

vector<int>e[N];
set<LL> s;
int A[N], B[N], deg[N], Link[N];
bool vis[N];

int main() {
    int n, m;
    while (~scanf("%d%d",&n, &m)) {
        int Size = sqrt(m);
        s.clear();
        for (int i = 1; i <= n; ++i) Link[i] = deg[i] = 0, vis[i] = false, e[i].clear();
        for (int i = 1; i <= m; ++i) {
            int u = read(), v = read();
            deg[u] ++, deg[v] ++;
            e[u].push_back(v), e[v].push_back(u);
            s.insert(1ll * u * n + v), s.insert(1ll * v * n + u);
        }
        LL Ans = 0;
        for (int u = 1; u <= n; ++u) {
            vis[u] = true; // 如果没有vis，Ans需要/2 
            for (int i = 0; i < e[u].size(); ++i) Link[e[u][i]] = u;
            for (int i = 0; i < e[u].size(); ++i) {
                int v = e[u][i], cnt = 0;
                if (vis[v]) continue;
                if (deg[v] <= Size) {
                    for (int j = 0; j < e[v].size(); ++j) 
                        if (Link[e[v][j]] == u) cnt ++;
                }
                else {
                    for (int j = 0; j < e[u].size(); ++j) 
                        if (s.find(1ll * e[u][j] * n + v) != s.end()) cnt ++;
                }
                Ans += (1ll * cnt * (cnt - 1) / 2);
            }
        }
        printf("%lld
",Ans);
    }
    return 0;
}