HDU6152-Friend-Graph

最新推荐文章于 2022-09-05 21:14:58 发布

Wang_128

最新推荐文章于 2022-09-05 21:14:58 发布

阅读量218

点赞数

CC 4.0 BY-SA版权

分类专栏： HDU ----其他

本文链接：https://blog.youkuaiyun.com/a664607530/article/details/77448544

HDU 同时被 2 个专栏收录

490 篇文章

订阅专栏

----其他

216 篇文章

订阅专栏

本文介绍了一种算法，用于判断一个由n个成员组成的团队是否为好团队。根据团队成员之间的朋友关系，若存在三个或以上的成员互为朋友或互不为友，则该团队被视为坏团队。文章提供了一个具体的实现案例，并解释了为何超过六人的团队不可能是好团队。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Friend-Graph

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1093 Accepted Submission(s): 569

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

题意：一个团队有n个人，如果有大于等于3个人相互之间是朋友或非朋友关系，那么这不是一个好团队

解题思路：其实大于等于6个人的团队必然不是一个好团队，那么小于6个人的团队暴力跑即可，所有总体跑大暴力也是可以过得，只是数据要用bool来记录

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int n, a, x[10][10];

int main()
{
    int t;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%d", &n);
        if (n >= 6)
        {
            for (int i = 1; i < n; i++)
                for (int j = i + 1; j <= n; j++)
                    scanf("%d", &a);
            printf("Bad Team!\n");
            continue;
        }
        memset(x, 0, sizeof x);
        for (int i = 1; i < n; i++)
        {
            x[i][i] = 0;
            for (int j = i + 1; j <= n; j++)
                scanf("%d", &a), x[i][j] = x[j][i] = a;
        }
        int flag = 1;
        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= n; j++)
            {
                if (i == j) continue;
                for (int k = 1; k <= n; k++)
                {
                    if (k == i || k == j) continue;
                    if ((x[i][j] && x[i][k] && x[j][k]) || (!x[i][j] && !x[i][k] && !x[j][k])) flag = 0;
                }
            }
        }
        if (flag) printf("Great Team!\n");
        else printf("Bad Team!\n");
    }
    return 0;
}