那這個轉移是什麼意思呢
就是列舉第$i$位置填的有多少人，設為$k$，然後組合數的意思是$sum[i]$個人中，肯定是有$cnt[i]$個人是不能選的（這些人已經強制選$i$了），然後再扣掉之前選的$j - k$個人，從中選$k - cnt[i]$個人

這樣就做完了

$Ps.$記得多組資料。。。。。。

程式碼：

#include<cstdio>
#include<vector>
#include<queue>
#include<ctime>
#include<algorithm>
#include<cstdlib>
#include<stack>
#include<cstring>
#include<cmath>
using namespace std;

typedef long long LL;

const int INF = 2147483647;
const int maxn = 310;

int n,m,crz;
int sum[maxn],cnt[maxn];
int f[maxn][maxn],C[maxn][maxn];

inline LL getint()
{
    LL ret = 0,f = 1;
    char c = getchar();
    while (c < '0' || c > '9')
    {
        if (c == '-') f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') ret = ret * 10 + c - '0',c = getchar();
    return ret * f;
}

inline int mod(int x)
{
    return x >= crz ? x - crz : x;
}

inline void init()
{
    C[0][0] = 1;
    for (int i = 1; i <= n; i++)
    {
        C[i][0] = 1;
        for (int j = 1; j <= i; j++)
            C[i][j] = mod(C[i - 1][j] + C[i - 1][j - 1]);
    }
}

inline void solve()
{
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            f[i][j] = 0;
    for (int i = 1; i <= n; i++) cnt[i] = 0;
    for (int i = 1; i <= n; i++) sum[i] = 0;

    n = getint(); m = getint(); crz = getint();

    init();

    for (int i = 1; i <= m; i++)
    {
        int p = getint(),q = getint();
        cnt[q]++;
    }

    sum[n] = n;
    for (int i = n - 1; i >= 1; i--)
        sum[i] = sum[i + 1] - cnt[i + 1];

    for (int i = 1; i <= n; i++)
        if (sum[i] < i) {printf("NO\n"); return;}

    printf("YES ");

    f[0][0] = 1;
    for (int i = 1; i <= n; i++)
        for (int j = i; j <= sum[i]; j++)
            for (int k = cnt[i]; k <= j - i + 1; k++)
                f[i][j] = mod(f[i][j] + 1ll * f[i - 1][j - k] * C[sum[i] - cnt[i] - (j - k)][k - cnt[i]] % crz);

    printf("%d\n",f[n][n]);
}

int main()
{
    #ifdef AMC
        freopen("AMC1.txt","r",stdin);
    #endif
    int t = getint();
    while (t--) solve();
    return 0;
}