Friend-Graph(数学，思维）

最新推荐文章于 2022-06-24 12:37:59 发布

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最新推荐文章于 2022-06-24 12:37:59 发布

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分类专栏：思维数学文章标签： ACM 思维数学

本文链接：https://blog.youkuaiyun.com/w571523631/article/details/77417243

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该博客介绍了一个团队友谊评估问题，其中定义了好团队的条件：不存在三人互为朋友或互为非朋友的情况。给出团队人数和成员间的关系，需要判断团队是否符合好团队的标准。博主提供了两种解决方案：一种基于拉姆齐定理，另一种通过查找三元环。代码实现包括两种方法，分别用C++编写，以解决此问题。

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Problem Description

It is well known that small groups are not conducive of the development of a team. Therefore, there shouldn’t be any small groups in a good team.
In a team with n members,if there are three or more members are not friends with each other or there are three or more members who are friends with each other. The team meeting the above conditions can be called a bad team.Otherwise,the team is a good team.
A company is going to make an assessment of each team in this company. We have known the team with n members and all the friend relationship among these n individuals. Please judge whether it is a good team.

Input

The first line of the input gives the number of test cases T; T test cases follow.（T<=15）
The first line od each case should contain one integers n, representing the number of people of the team.(

n≤3000 )

Then there are n-1 rows. The

i th row should contain n-i numbers, in which number

aij represents the relationship between member i and member j+i. 0 means these two individuals are not friends. 1 means these two individuals are friends.

Output

Please output ”Great Team!” if this team is a good team, otherwise please output “Bad Team!”.

Sample Input

Sample Output

  
  
   
   Great Team!

Source

2017中国大学生程序设计竞赛 - 网络选拔赛

题目思路：

这题可以用两种方法来写，第一种因为实际上是一个完全图，所以根据拉姆齐定理，大于等于6的直接输出no，剩下的暴力找三元环。

第二种根据题目意思很容易知道是找三元环了，对于每一个点我们去找符合题意的三元环，就是找一个有边的，找一个没有边的，最后去下重，然后判断是不是等于全部三元环的个数就行了。

第一种：

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cstring>
#include<iostream>
#include<sstream>
#include<cmath>
#include<vector>
#define LL long long
#define INF 0x3f3f3f3f
#define eps 1e-6
using namespace std;
using namespace std;
const LL mod = 1e9 + 7;
int mp[10][10];
int n;
int T;
int main()
{
    scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        memset(mp,0,sizeof(mp));
        if(n>=6){
        for(int i = 1;i<=n;i++){
            for(int j = 1;j<=n-i;j++){
                int x;
                scanf("%d",&x);
            }
        }
        }
        else{
            for(int i = 1;i<=n;i++){
            for(int j = 1;j<=n-i;j++){
                int x;
                scanf("%d",&x);
                mp[i][i+j] =x;
                mp[i+j][i] = x;
            }
        }
        }
        if(n>=6){
            puts("Bad Team!");
        }
        else{
                int falg = 1;
            for(int i = 1;i<=n;i++){
                for(int j = 1;j<=n;j++){
                    for(int k = 1;k<=n;k++){
                        if(i==j||j==k||i==k)
                            continue;
                        if(mp[i][j]==1&&mp[i][k]==1&&mp[j][k]==1){
                            falg= 0;
                            //cout<<"fuck1"<<endl;
                        }
                        if(mp[i][j]==0&&mp[i][k]==0&&mp[j][k]==0){
                            //cout<<i<<' '<<j<<' '<<k<<endl;
                            //cout<<"fuck2"<<endl;
                            falg = 0;
                        }
                    }
                }
            }
            if(falg)
                puts("Great Team!");
            else
                puts("Bad Team!");
        }
    }
}

第二种：

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<memory.h>
#include<iostream>
#define eps 1e-6
#define LL long long
using namespace std;
int cnt[3005];
int n;
int T;
int main()
{
    scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        memset(cnt,0,sizeof(cnt));
        for(int i = 1;i<=n;i++){
            for(int j = 1;j<=n-i;j++){
                int x;
                scanf("%d",&x);
                if(x){
                    cnt[i]++;
                    cnt[i+j]++;
                }
            }
        }
        LL sum = (LL)n*(n-1)*(n-2)/6;
        LL sum2 = 0;
        for(int i = 1;i<=n;i++){
            sum2+=(LL)(cnt[i])*(n-cnt[i]-1);
        }
        if(sum2/2==sum)
            puts("Great Team!");
        else
            puts("Bad Team!");
    }
}