题目链接:这里
题意:一个n*m的图,让你从左上角走到右下角,有一些点不能经过,问你有多少种方法
解法:令dp[i]表示从原点不经过任何坏点到第i个点的方案数
这里:这里写链接内容
//CF 559C
#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9+7;
const int maxn = 1000010;
#define ll long long
struct Point{
ll x, y;
Point(){}
Point(ll x, ll y) : x(x), y(y) {}
bool operator <(const Point &rhs) const{
if(x == rhs.x) return y < rhs.y;
return x < rhs.x;
}
}p[5000];
ll fac[maxn];
ll powmod(ll a, ll b){
ll res = 1;
while(b){
if(b&1) res = res*a%mod;
a = a*a%mod;
b>>=1;
}
return res;
}
ll C(ll n, ll m){
if(m > n || m < 0) return 0;
ll s1 = fac[n], s2 = fac[m]*fac[n-m]%mod;
return s1*powmod(s2, mod-2)%mod;
}
ll dp[maxn];
int n, m, k;
int main(){
fac[0] = 1LL;
for(int i = 1; i < maxn; i++) fac[i] = fac[i-1] * i % mod;
scanf("%d%d%d", &n, &m, &k);
for(int i = 1; i <= k; i++){
scanf("%lld%lld", &p[i].x, &p[i].y);
p[i].x--, p[i].y--;
}
p[++k].x = n-1;
p[k].y = m-1;
sort(p+1, p+k+1);
for(int i = 1; i <= k; i++){
dp[i] = C(p[i].x + p[i].y, p[i].x);
for(int j = 1; j < i; j++){
if(p[j].y <= p[i].y){
dp[i] += (mod - dp[j]*C(p[i].x-p[j].x+p[i].y-p[j].y, p[i].x-p[j].x));
dp[i] %= mod;
}
}
}
printf("%lld\n", (dp[k]+mod)%mod);
}