hdu5852 Intersection is not allowed! 【矩阵行列式】

题意##

给出\(n*n\)网格\((n<=10^5)\)

顶部有\(K\)个起点，底部有\(K\)个相对应的终点

每次只能向下或向右走

求有多少种从各个起点出发到达对应终点且路径不相交的路径？

对\(10^9 + 7\)取模

题解

按照组合数的套路

二维空间从一个h * w的矩形左上角走到右下角只向下或向右走的方案数为\(C_{h + w}^{h}\)

这题直接求不好求，我们考虑容斥

可以发现，如果两个路径相交，就相当于交换了终点

那么我们枚举每个起点的终点，乘上一个容斥系数\((-1)^t\)，其中\(t\)表示相交的路径个数，也可以看做选择的终点序列的逆序对个数

按照每个起点择每一个终点构造出一个组合数的矩阵，按照矩阵的的定义你会发现其实答案就等于这个矩阵的行列式

我们求一个行列式就做完了

这题太神了

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define LL long long int
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define BUG(s,n) for (int i = 1; i <= (n); i++) cout<<s[i]<<' '; puts("");
using namespace std;
const int maxn = 105,maxm = 200005,INF = 1000000000,P = 1000000007;
inline int read(){
	int out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
	return out * flag;
}
LL fac[maxm],inv[maxm],fv[maxm];
void init(){
	fac[0] = 1;
	for (int i = 1; i <= 200000; i++) fac[i] = fac[i - 1] * i % P;
	inv[0] = inv[1] = 1;
	for (int i = 2; i <= 200000; i++) inv[i] = (P - P / i) * inv[P % i] % P;
	fv[0] = 1;
	for (int i = 1; i <= 200000; i++) fv[i] = fv[i - 1] * inv[i] % P;
}
LL qpow(LL a,LL b){
	LL ans = 1;
	for (; b; b >>= 1,a = a * a % P)
		if (b & 1) ans = ans * a % P;
	return ans;
}
LL C(int n,int m){
	if (m > n) return 0;
	return fac[n] * fv[m] % P * fv[n - m] % P;
}
int n,K,a[maxn],b[maxn];
LL A[maxn][maxn],ans;
void gause(){
	ans = 1;
	for (int i = 1; i <= K; i++){
		int pos = i;
		while (A[pos][i] == 0 && pos <= K) pos++;
		if (pos > K){ans = 0; return;}
		if (pos != i) for (int j = i; j <= K; j++) swap(A[i][j],A[pos][j]),ans = -ans;
		for (int j = i + 1; j <= K; j++){
			LL t = A[j][i] * qpow(A[i][i],P - 2) % P;
			for (int k = i; k <= K; k++)
				A[j][k] = (A[j][k] - A[i][k] * t % P + P) % P;
		}
		ans = ans * A[i][i] % P;
	}
	ans = (ans % P + P) % P;
}
int main(){
	ios::sync_with_stdio(false);
	init();
	int T = read();
	while (T--){
		n = read(); K = read();
		REP(i,K) a[i] = read();
		REP(i,K) b[i] = read();
		REP(i,K) REP(j,K) A[i][j] = C(n - 1 + b[j] - a[i],n - 1);
		gause();
		cout << ans << endl;
	}
	return 0;
}