There are N sports players. Among them, there are M incompatible pairs. The i-th incompatible pair (1≤i≤M) is the A i ​ -th and B i ​ -th players. You will divide the players into T teams. Every player must belong to exactly one team, and every team must have one or more players. Additionally, for each i=1,2,…,M, the A i ​ -th and B i ​ -th players must not belong to the same team. Find the number of ways to satisfy these conditions. Here, two divisions are considered different when there are two players who belong to the same team in one division and different teams in the other.c++代码

以下是一种C++代码实现的方式：

#include <vector>

using namespace std;

int N, M, T;
vector<int> A, B;
vector<int> team;

int dfs(int idx) {
    if (idx == N) {
        return 1;
    }
    
    int res = 0;
    for (int i = 1; i <= T; i++) {
        bool valid = true;
        for (int j = 0; j < M; j++) {
            if ((A[j] == idx && team[B[j]] == i) || (B[j] == idx && team[A[j]] == i)) {
                valid = false;
                break;
            }
        }
        
        if (valid) {
            team[idx] = i;
            res += dfs(idx + 1);
            team[idx] = 0;
        }
    }
    
    return res;
}

int main() {
    cin >> N >> M >> T;
    
    A.resize(M);
    B.resize(M);
    team.resize(N);
    
    for (int i = 0; i < M; i++) {
        cin >> A[i] >> B[i];
    }
    
    int ans = dfs(0);
    cout << ans << endl;
    
    return 0;
}

该代码使用深度优先搜索（DFS）的方法来枚举所有可能的分组方式，并通过检查每对不兼容的球员是否在同一队来验证分组合法性。最终输出满足条件的分组数目。

注意：这只是一种实现方式，具体解决方案可能有多种，可以根据实际情况进行适当调整。