HDU 6152.Friend-Graph




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 $i$th row should contain n-i numbers, in which number $a_{i j}$ 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 1 0
0 0

Great Team!



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)


#include <cstdio>
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)
        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++)
            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;
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