cf559C. Gerald and Giant Chess(容斥原理)

题意

$h \times w$的网格，有$n$个障碍点，

每次可以向右或向下移动

求从$(1, 1)$到$(h, w)$不经过障碍点的方案数

sol

容斥原理

从$(1, 1)$到$(h, w)$不经过障碍点的方案数为$c(h + w, h)$

设$f[i]$表示到达第$i$个黑格子的合法路径的方案数

首先对所有点按$x$排序，这样就能保证每次从他的左上方转移而来

然后根据公式算一下就好了

// luogu-judger-enable-o2
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<stack>
#include<vector>
#include<cstring>
#define pair pair<int, int>
#define mp(x, y) make_pair(x, y)
#define fi first
#define se second
#define int long long 
//#define int long long
using namespace std;
const int maxn = 3 * 1e6, mod = 1e9 + 7;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9'){if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
pair p[maxn];
int h, w, n;
int fac[maxn], ifac[maxn], f[maxn];
int fastpow(int a, int p) {
    int base = 1;
    while(p) {
        if(p & 1) base = (base * a) % mod;
        a = (a * a) % mod; p >>= 1;
    }
    return base % mod;
}
int c(int n, int m) {
    return (fac[n] * ifac[m] % mod * ifac[n - m]) % mod;
}
main() {
    h = read(), w = read(); n = read() + 1; 
    fac[0] = 1; for(int i = 1; i <= h + w; i++) fac[i] = i * fac[i - 1] % mod;
    ifac[h + w] = fastpow(fac[h + w], mod - 2); 
    for(int i = h + w; i >= 1; i--) ifac[i - 1] = i * ifac[i] % mod;
    for(int i = 1; i <= n - 1; i++) {
        int x = read(), y = read();
        p[i] = mp(x, y);
    }
    p[n] = mp(h, w);
    sort(p + 1, p + n + 1);
    for(int i = 1; i <= n; i++) {
        f[i] = c(p[i].fi + p[i].se - 2, p[i].fi - 1);
        for(int j = i - 1; j >= 1; j--) {
            if(p[j].se <= p[i].se) {
                int x = p[i].fi - p[j].fi + 1, y = p[i].se - p[j].se + 1;
                (f[i] -= f[j] * c(x + y - 2, x - 1) % mod + mod) %= mod;
            }
        }
    }
    printf("%i64d", (f[n] + mod) % mod);
    return 0;
}
/*
2 3 2
2 1
2 2
*/