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题目
{% fold 点击显/隐题目 %}
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.
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.($n \leq 3000$)
Then there are n-1 rows. The iith row should contain n-i numbers, in which number aija_{i j} 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.
Please output ”Great Team!” if this team is a good team, otherwise please output “Bad Team!”.
1 4 1 1 0 0 0 1
Great Team!
题解
判断一个无向图中是否存在3个点没有直接联通，或者3个点直接互联通
可以转换为是否有点有三个及三个以上的边,或者有三个以上的点与他没有边
可以发现5个点按照正五边形连接是极端情况了,再多一个点,无论怎么连都无法满足要求
因此,当 n>=6 时必定是 Bad Team!
对于 n<6 的情况暴力求解即可
需要注意的是,数组开到 7*7 即可,开 1e5*1e5 会爆内存
另外如果 n>=6 ,读入仍然需要一个一个读,不能一次读一行(最后一行没有 \n)
代码
{% fold 点击显/隐代码 %}```cpp Friend-Graph https://github.com/OhYee/sourcecode/tree/master/ACM 代码备份
#include
using namespace std;
const int maxn = 10;
int n;
int r[maxn][maxn];
bool judge() {
for (int i = 1; i <= n; ++i) {
int friends = 0, notfriends = 0;
for (int j = 1; j <= n; ++j) {
if (i != j) {
if (r[i][j] == 1)
++friends;
else
++notfriends;
}
}
if (friends >= 3 || notfriends >= 3)
return false;
}
return true;
}
int main() {
int T;
scanf("%d", &T);
while (T--) {
scanf("%d", &n);
bool ok;
if (n >= 6) {
for (int i = 1; i < n; i++)
scanf("%*[^\n]");
ok = false;
} else {
for (int i = 1; i < n; i++)
for (int j = 1; j <= n - i; ++j) {
scanf("%d", &r[i][i + j]);
r[i + j][i] = r[i][i + j];
}
ok = judge();
}
printf("%s\n", ok ? "Great Team!" : "Bad Team!");
}
return 0;
}
{% endfold %}