hdu 6184 三元环数目

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Counting Stars

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 358    Accepted Submission(s): 90

Problem Description

Little A is an astronomy lover, and he has found that the sky was so beautiful!

So he is counting stars now!

There are n stars in the sky, and little A has connected them by m non-directional edges.

It is guranteed that no edges connect one star with itself, and every two edges connect different pairs of stars.

Now little A wants to know that how many different "A-Structure"s are there in the sky, can you help him?

An "A-structure" can be seen as a non-directional subgraph G, with a set of four nodes V and a set of five edges E.

If V=(A,B,C,D) and E=(AB,BC,CD,DA,AC), we call G as an "A-structure".

It is defined that "A-structure" G1=V1+E1 and G2=V2+E2 are same only in the condition that V1=V2 and E1=E2.

 

Input

There are no more than 300 test cases.

For each test case, there are 2 positive integers n and m in the first line.

2≤n≤105, 1≤m≤min(2×105,n(n−1)2)

And then m lines follow, in each line there are two positive integers u and v, describing that this edge connects node u and node v.

1≤u,v≤n

∑n≤3×105,∑m≤6×105

 

Output

For each test case, just output one integer--the number of different "A-structure"s in one line.

 

Sample Input

4 51 22 33 44 11 34 61 22 33 44 11 32 4

 

Sample Output

16

 

Source

2017ACM/ICPC广西邀请赛-重现赛（感谢广西大学）

 

Recommend

liuyiding

题意：规定V=(A,B,C,D) and E=(AB,BC,CD,DA,AC)这种图算一种。给你一副图问有多少种。

分析：仔细分析一张这种图其实是两个三元环共用一条边得到，如果我能计算对边有多少条，那么我就能得出最后的答案。

我们考虑在求出三元环的过程中标记一下每条边有几个对点，最后算一下总数即可。

//china no.1#pragma comment(linker, "/STACK:1024000000,1024000000")#include <vector>#include <iostream>#include <string>#include <map>#include <stack>#include <cstring>#include <queue>#include <list>#include <stdio.h>#include <set>#include <algorithm>#include <cstdlib>#include <cmath>#include <iomanip>#include <cctype>#include <sstream>#include <functional>#include <stdlib.h>#include <time.h>#include <bitset>using namespace std;#define pi acos(-1)#define PI acos(-1)#define endl '\n'#define srand() srand(time(0));#define me(x,y) memset(x,y,sizeof(x));#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)#define close() ios::sync_with_stdio(0); cin.tie(0);#define FOR(x,n,i) for(int i=x;i<=n;i++)#define FOr(x,n,i) for(int i=x;i<n;i++)#define fOR(n,x,i) for(int i=n;i>=x;i--)#define fOr(n,x,i) for(int i=n;i>x;i--)#define W while#define sgn(x) ((x) < 0 ? -1 : (x) > 0)#define bug printf("***********\n");#define db double#define ll long long#define mp make_pair#define pb push_backtypedef long long LL;const int INF=0x3f3f3f3f;const LL LINF=0x3f3f3f3f3f3f3f3fLL;const int dx[]={-1,0,1,0,1,-1,-1,1};const int dy[]={0,1,0,-1,-1,1,-1,1};const int maxn=1e3+10;const int maxx=1e5+100;const double EPS=1e-8;const double eps=1e-8;const int mod=1e9+7;template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}template <class T>inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}if(IsN) num=-num;return true;}void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}void print(LL a){ Out(a),puts("");}//freopen( "in.txt" , "r" , stdin );//freopen( "data.txt" , "w" , stdout );//cerr << "run time is " << clock() << endl;int n,m;vector<int>G[maxx];int vis[maxx],link[maxx];int main(){    W(~scanf("%d%d",&n,&m))    {        int u,v;        me(vis,0);        me(link,-1);        int limit=ceil(sqrt(m));        FOR(1,m,i)        {            scan_d(u),scan_d(v);            G[u].pb(v);            G[v].pb(u);        }        FOR(1,n,i)            sort(G[i].begin(),G[i].end());        LL ans=0;        FOR(1,n,i)        {            vis[i]=1;            for(int y:G[i]) link[y]=i;            for(int y:G[i])            {                if(vis[y]) continue;                LL cnt=0;                if(G[y].size()<=limit)                {                    for(int z:G[y])                    {                        if(link[z]==i) cnt++;                    }                }                else                {                    for(int z:G[i])                    {                        if(z!=y,binary_search(G[z].begin(),                            G[z].end(),y))                            cnt++;                    }                }                ans+=cnt*(cnt-1)/2;            }        }        print(ans);        FOR(1,n,i)            G[i].clear();    }}

E=(AB,BC,CD,DA,AC)E=(AB,BC,CD,DA,AC)E=(AB,BC,CD,DA,AC)E=(AB,BC,CD,DA,AC)

E=(AB,BC,CD,DA,AC)问我饿