【JZOJ6403】【NOIP2019模拟11.04】a

题目大意

从\((0,0,0)\)走到\((n,n,n)\)，每一步可以从\((x,y,z)\)走到\((x+1,y,z)\)或\((x,y+1,z)\)或\((x,y,z+1)\)，给出\(m\)个不允许经过的点，求这样走的方案数。
\(n\leq 100000,m\leq 5000\)

Solution

很经典的题目。

枚举\(i\)个点必定经过然后容斥，复杂度\(O(m!)\)。

考虑把这个过程放到递推上，设\(f_i\)表示从\((0,0,0)\)走到第\(i\)个点，且不经过其他点的方案数；记\(j \to i\)表示\(j\)能到达\(i\)，\(g(a,b,c)\)表示从\((0,0,0)\)以任意方式走到\((a,b,c)\)的方案数，有：
\[f_i=\sum_{j \to i}-f_j*g(x_i-x_j,y_i-y_j,z_i-z_j)\]
怎么推的？\(f_j\)里的所有路径经过\(i\)后都多了一个点，所以容斥系数都要乘以\(-1\)。

这样就做到了\(O(m^2)\)。

Code

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

typedef long long ll;
const int N = 300017, M = 5017;
const ll P = 1000000007;

int n, m, tot;
ll fac[N], inv[N], f[M];
struct note {
    int x, y, z;
} a[M];
int cmp(note p, note q) {
    return p.x == q.x ? (p.y == q.y ? p.z < q.z : p.y < q.y) : p.x < q.x;
}
int operator==(note p, note q) { return p.x == q.x && p.y == q.y && p.z == q.z; }

void pre() {
    fac[0] = 1;
    for (int i = 1; i <= 300000; ++i) fac[i] = fac[i - 1] * i % P;
    inv[0] = inv[1] = 1;
    for (int i = 2; i <= 300000; ++i) inv[i] = inv[P % i] * (P - P / i) % P;
    for (int i = 2; i <= 300000; ++i) inv[i] = inv[i] * inv[i - 1] % P;
}
ll C(int n, int m) {
    return fac[n] * inv[m] % P * inv[n - m] % P;
}
ll calc(int dx, int dy, int dz) {
    return C(dx + dy + dz, dx) * C(dy + dz, dy) % P;
}

int main() {
    //freopen("a.in", "r", stdin);
    //freopen("a.out", "w", stdout);
    pre();
    scanf("%d%d", &n, &m);
    a[++tot] = (note){0, 0, 0};
    for (int i = 1; i <= m; ++i) ++tot, scanf("%d%d%d", &a[tot].x, &a[tot].y, &a[tot].z);
    a[++tot] = (note){n, n, n};
    sort(a + 1, a + tot + 1, cmp);
    tot = unique(a + 1, a + tot + 1) - a - 1;
    f[1] = P - 1;
    for (int i = 2; i <= tot; ++i)
        for (int j = 1; j < i; ++j)
            if (a[j].x <= a[i].x && a[j].y <= a[i].y && a[j].z <= a[i].z)
                f[i] = (f[i] - calc(a[i].x - a[j].x, a[i].y - a[j].y, a[i].z - a[j].z) * f[j] % P + P) % P;
    printf("%lld\n", f[tot]);
    return 0;
}