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cf559C. Gerald and Giant Chess(容斥原理)

2018-09-05 07:43:16来源：博客园阅读 ()

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
*/