Friend-Graph(HDU 3152)
来源:互联网 发布:动物相机软件 编辑:程序博客网 时间:2024/06/04 23:30
Friend-Graph
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1752 Accepted Submission(s): 870
Total Submission(s): 1752 Accepted Submission(s): 870
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.
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. Thei 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.
The first line od each case should contain one integers n, representing the number of people of the team.(
Then there are n-1 rows. The
Output
Please output ”Great Team!” if this team is a good team, otherwise please output “Bad Team!”.
Sample Input
141 1 00 01
Sample Output
Great Team!
//题目:有3个人以上互相认识或互相不认识就是Bad Team,不然就是Good Team。给定一张图,求Good Team/Bad Team。
//思路:一开始用最大团做,不出意外的TLE了。后来发现暴力搜整张图中是否存在节点个数为3的环即可,图正反各求一遍。PS:只能用一个二维数组,否则会MLE。
#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <cmath>#include <vector>#include <queue>#include <algorithm>using namespace std;const short MAX=3005;int n;bool map[MAX][MAX];bool vis[MAX];//搜是否存在大小为3的环bool judge(){ memset(vis,false,sizeof(vis)); int i,j; queue<int>q; q.push(1); vis[1]=true; while(!q.empty()) { int now=q.front(); q.pop(); for(i=1;i<n;i++) { for(j=i+1;j<=n;j++) { if(map[i][j]==true&&map[now][i]==true&&map[now][j]==true) { return false; } } if(map[i][j]==true&&vis[i]==false) { vis[i]=true; q.push(i); } } } return true;}int main(){ int T; scanf("%d",&T); while(T--) { scanf("%d",&n); for(int i=1;i<n;i++) { for(int j=i+1;j<=n;j++) { int num; scanf("%d",&num); if(num==1) map[i][j]=map[j][i]=true; else map[i][j]=map[j][i]=false; } } if(judge()==false) { printf("Bad Team!\n"); continue; } //建反图 for(int i=1;i<n;i++) { for(int j=i+1;j<=n;j++) { if(map[i][j]==true) { map[i][j]=map[j][i]=false; } else { map[i][j]=map[j][i]=true; } } } if(judge()==false) printf("Bad Team!\n"); else printf("Great Team!\n"); } return 0;}
阅读全文
0 0
- Friend-Graph(HDU 3152)
- HDU Friend-Graph(离散数学)
- HDU 6152-Friend-Graph(ccpc)
- HDU 6152 Friend-Graph(定理)
- Hdu 6152 Friend-Graph(Ramsey定理)
- HDU 6152 Friend-Graph (最大团)
- HDU 6152 Friend-Graph(暴力)
- HDU-6152 Friend-Graph (思维+暴力)
- hdu 6152 Friend-Graph(ccpc)
- HDU 6152 Friend-Graph (CCPC2017)
- hdu 6152 Friend-Graph(拉姆齐定理)
- HDU 6152 Friend-Graph
- HDU 6152 Friend-Graph
- hdu-6152 Friend-Graph
- HDU 6152 Friend-Graph
- HDU 6152 Friend-Graph
- hdu 6152 Friend-Graph
- Friend-Graph HDU
- tcp keeplive
- 从Uncaught SyntaxError: Unexpected token ")" 问题看javascript:void的作用
- HDU 6141 I am your Father!(最小树形图)
- vue项目中浏览器图标的设置
- bjui关闭当前页及弹窗层,刷新当前页及弹窗层
- Friend-Graph(HDU 3152)
- Qt 学习之路 2(36):二进制文件读写
- 详情页返回列表内容缓存及定位实现
- 去掉alert,confirm弹出框显示的url
- poi操作word文档(替换,插入图片)
- 剑指Offer 面试题14 剪绳子
- 发送手机验证码在activeMQ中简单应用
- 状态真的特别重要
- Python引用其他模块之sys.path