Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

BZOJ 4767. 两双手

#4767. 两双手

先将坐标变换为以 $A, B$ 为基底的坐标,这样两种走法分别为向上、向右走一格。这样坐标的范围到了 $500 \times 500$ ,不能直接 DP 完成。考虑容斥,$f _ i$ 表示第一个到达的关键点的方案数。记 $g _ {i, j}$ 表示从第 $i$ 个点到第 $j$ 个点没有任何限制的方案数,可以用组合数简单地算出,那么有 $f _ i = g _ {0, i} - \sum _ {x _ j \le x _ i \wedge y _ j \le y _ i} g _ {j, i} f _ j$ 。注意将重复的关键点去除。

查看代码
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#include <cstdio>
#include <utility>
#define fi first
#define se second
using namespace std;
template <class Type>
void read(Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
c == '-' && (flag = true);
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 3) + (x << 1) + c - '0';
if (flag) x = ~x + 1;
}
template <class Type, class ...rest>
void read(Type &x, rest &...y) { read(x), read(y...); }
template <class Type>
void write(Type x)
{
if (x < 0) putchar('-'), x = ~x + 1;
if (x > 9) write(x / 10);
putchar(x % 10 + '0');
}
typedef long long LL;
typedef pair <int, int> PII;
const int N = 5e5 + 10, M = 510, mod = 1e9 + 7;
PII p[M];
bool vis[M];
int fact[N], ifact[N];
int n, Ax, Ay, Bx, By, f[M];
int binpow (int b, int k = mod - 2)
{
int res = 1;
for (; k; k >>= 1, b = (LL)b * b % mod)
if (k & 1) res = (LL)res * b % mod;
return res;
}
void init ()
{
fact[0] = 1;
for (int i = 1; i < N; ++i)
fact[i] = (LL)fact[i - 1] * i % mod;
ifact[N - 1] = binpow(fact[N - 1]);
for (int i = N - 1; i; --i)
ifact[i - 1] = (LL)ifact[i] * i % mod;
}
int calc (int a, int b) { return (LL)ifact[a] * ifact[b] % mod * fact[a + b] % mod; }
int dfs (int x)
{
if (vis[x]) return f[x];
vis[x] = true;
f[x] = calc(p[x].fi, p[x].se);
for (int i = 0; i <= n; ++i)
if (p[i].fi <= p[x].fi && p[i].se <= p[x].se && p[i] != p[x])
f[x] = (f[x] - (LL)dfs(i) * calc(p[x].fi - p[i].fi, p[x].se - p[i].se)) % mod;
return f[x];
}
int main ()
{
init();
read(p[0].fi, p[0].se, n, Ax, Ay, Bx, By);
for (int i = 1; i <= n; ++i) read(p[i].fi, p[i].se);
int D = Ax * By - Bx * Ay;
for (int i = 0; i <= n; ++i)
{
int D1 = p[i].fi * By - Bx * p[i].se, D2 = Ax * p[i].se - p[i].fi * Ay;
if (D1 % D || D2 % D) vis[i] = true;
p[i].fi = D1 / D, p[i].se = D2 / D;
if (p[i].fi < 0 || p[i].se < 0) vis[i] = true;
for (int j = 0; j < i && !vis[i]; ++j) if (p[i] == p[j]) vis[i] = true;
}
write((dfs(0) + mod) % mod);
return 0;
}