路径计数2（dp + 组合数）

daimayuan 每日一题 div1 202

Solution

1. dp[i] 表示走到第i障碍物的合法方案数
2. 走到点（x， y）的所有方案数：$C(x+y-2，x-1)$
3. $dp[i]=C(i.x+i.y-2,i.x-1)-\sum dp[j]\ast C(i.x-j.x+i.y-j.y,i.x-j.x)$
   即：合法方案数 = 所有方案数 - 所有非法的方案数

Code

const ll mod = 1000000007;
using namespace std;
const int N = 2e6 + 5;

ll f[N];
ll inv[N], fac[N];
void init()
{
    
	fac[0] = inv[1] = inv[0] = 1;
	//注意：inv[0] = 1, 即零的阶乘的逆元为1
	for (int i = 1; i <= 2e6; i++)
		fac[i] = fac[i - 1] * i % mod;
	
	//先求出1~n的逆元
	for (int i = 2; i <= 2e6; i++)
		inv[i] = (mod - mod / i) * inv[mod % i] % mod;
	//1~n的逆元累成即为 1~n的阶乘的逆元
	for (int i = 2; i <= 2e6; i++)
		inv[i] = inv[i] * inv[i - 1] % mod;
}

ll C(int n, int m)
{
    
	return fac[n] * inv[n - m] % mod * inv[m] % mod;
}

pii s[N];
int main()
{
    
	IOS;
	init();
	int n, m; cin >> n >> m;
	for (int i = 1; i <= m; i++)
		cin >> s[i].ft >> s[i].sd;

	s[m + 1] = {
     n, n };
	for (int i = 1; i <= m + 1; i++)
	{
    
		f[i] = C(s[i].ft + s[i].sd - 2, s[i].ft - 1);
		for (int j = 1; j < i; j++)
		{
    
			ll t = C(s[i].ft - s[j].ft + s[i].sd - s[j].sd, s[i].ft - s[j].ft);
			f[i] = ((f[i] - f[j] * t % mod) % mod + mod) % mod;
		}
	}
	cout << f[m + 1] << endl;


	return 0;
}