BZOJ4031 [HEOI2015]小Z的房间 【矩阵树定理 + 高斯消元】

题目链接

BZOJ4031

题解

第一眼：这不裸的矩阵树定理么

第二眼：这个模\(10^9\)是什么鬼嘛QAQ

想尝试递归求行列式，发现这是\(O(n!)\)的。。

想上高斯消元，却又处理不了逆元这个东西、、

无奈去翻题解，，，

发现可以用类似辗转相除法去消，而避免除法，，，

这样子依旧是每次一行减去另一行乘一个数，行列式不变

orz

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#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 mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 105,maxm = 100005,INF = 1000000000,P = 1000000000;
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;
}
int visx[maxn],A[maxn][maxn],n,m,N,id[maxn][maxn];
int X[4] = {0,0,1,-1},Y[4] = {-1,1,0,0};
char s[maxn][maxn];
int Gause(){
	int tag = 1;
	for (int i = 1; i < N; i++){
		int j = i;
		for (int k = i + 1; k < N; k++)
			if (abs(A[k][i]) > abs(A[j][i])) j = k;
		if (j != i){
			for (int k = i; k < N; k++) swap(A[i][k],A[j][k]);
			tag = -tag;
		}
		for (j = i + 1; j < N; j++){
			while (A[j][i]){
				int tmp = A[j][i] / A[i][i];
				for (int k = i; k < N; k++)
					A[j][k] = ((A[j][k] - 1ll * tmp * A[i][k] % P) % P + P) % P;
				if (A[j][i]){
					for (int k = i; k < N; k++)
						swap(A[i][k],A[j][k]);
					tag = -tag;
				}
			}
		}
	}
	int re = 1;
	for (int i = 1; i < N; i++)
		re = 1ll * re * A[i][i] % P;
	return (re * tag % P + P) % P;
}
int main(){
	n = read(); m = read();
	REP(i,n){
		scanf("%s",s[i] + 1);
		REP(j,m) if (s[i][j] == '.') id[i][j] = ++N;
	}
	REP(i,n) REP(j,m){
		if (s[i][j] == '*') continue;
		int nx,ny,u = id[i][j];
		for (int k = 0; k < 4; k++){
			nx = i + X[k];
			ny = j + Y[k];
			if (nx < 1 || ny < 1 || nx > n || ny > m || s[nx][ny] == '*') continue;
			A[u][u]++;
			A[u][id[nx][ny]] = -1;
		}
	}
	printf("%d\n",Gause());
	return 0;
}