《算法竞赛进阶指南》0x5C计数DP Gerald & Giant Chess

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const int N = 2010,M=200010;
const int mod = 1000000007;
typedef long long ll;
#define PII pair<int,int>
struct node{
    int x,y;
    bool operator < (const node &a)const {
        if(x==a.x)return y<=a.y;
        else return x<=a.x;
    }
}a[N];
int f[N];
ll jc[M],jcinv[M];
int h,w,n;
ll ksm(ll a,int k){
    ll ans=1;
    while(k){
        if(k&1)ans=ans*a %mod;
        a=a*a %mod;
        k>>=1;
    }
    return ans;
}
int C(int n,int m){//%运算和* /是同一个优先级 
    return jc[n]*jcinv[m]%mod * jcinv[n-m]%mod;
}
int main(){
    cin>>h>>w>>n;
    for(int i=1;i<=n;i++){
        scanf("%d%d",&a[i].x,&a[i].y);
    }
    sort(a+1,a+n+1);
    a[n+1].x=h,a[n+1].y=w;
    a[0].x=a[0].y=1;
    
    jc[0]=1;
    jcinv[0]=1;
    for(int i=1;i<=200000;i++){//预处理阶乘逆元和阶乘 
        jc[i]=jc[i-1]*i %mod;
        jcinv[i]=ksm(jc[i],mod-2);
    }
    
    f[0]=1;
    for(int i=1;i<=n+1;i++){
        f[i]=C(a[i].x-1+a[i].y-1,a[i].x-1);
        for(int j=1;j<i;j++){
            if(a[j].x>a[i].x || a[j].y>a[i].y)continue;
            f[i] = (f[i] - (ll)f[j] * C(a[i].x + a[i].y - a[j].x - a[j].y, a[i].x - a[j].x)) % mod;
        }
    }
    
    cout<<(f[n+1]+mod)%mod<<endl;
    
    return 0;
}