URAL-1627-Join 生成树计数

传送门：https://vjudge.net/problem/URAL-1627

题意：

　　给定一个n*m的图，问图中“.”的点生成的最小生成树有多少个。

思路：

　　生成树的计数，需要用Kirchhoff矩阵。

　

实际中只开了一个矩阵，如果有一条边（u,v），那么把a[u][v]=a[v][u] = -1, a[u][u]++, a[v][v]++;

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>

using namespace std;
//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
// #pragma GCC diagnostic error "-std=c++11"
// #pragma comment(linker, "/stack:200000000")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
// #pragma GCC optimize("-fdelete-null-pointer-checks,inline-functions-called-once,-funsafe-loop-optimizations,-fexpensive-optimizations,-foptimize-sibling-calls,-ftree-switch-conversion,-finline-small-functions,inline-small-functions,-frerun-cse-after-loop,-fhoist-adjacent-loads,-findirect-inlining,-freorder-functions,no-stack-protector,-fpartial-inlining,-fsched-interblock,-fcse-follow-jumps,-fcse-skip-blocks,-falign-functions,-fstrict-overflow,-fstrict-aliasing,-fschedule-insns2,-ftree-tail-merge,inline-functions,-fschedule-insns,-freorder-blocks,-fwhole-program,-funroll-loops,-fthread-jumps,-fcrossjumping,-fcaller-saves,-fdevirtualize,-falign-labels,-falign-loops,-falign-jumps,unroll-loops,-fsched-spec,-ffast-math,Ofast,inline,-fgcse,-fgcse-lm,-fipa-sra,-ftree-pre,-ftree-vrp,-fpeephole2",3)

#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "
";
#define pb push_back
#define pq priority_queue



typedef long long ll;
typedef unsigned long long ull;

typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '
'

#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
#define max3(a,b,c) max(max(a,b), c); 
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;       
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e9;
const double esp = 1e-8;
const double PI=acos(-1.0);
const double PHI=0.61803399;    //黄金分割点
const double tPHI=0.38196601;


template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x=f?-x:x;
}


/*-----------------------showtime----------------------*/
            const int maxn = 300;
            char str[20];
            ll a[maxn][maxn];
            int g[maxn][maxn];
            int n,m,k;
            void cal(){
                ll ans = 1;int sign = 0;
                for(int i=1; i<=n; i++){        //当前行
                    for(int j=i+1; j<=n; j++){
                        int x = i, y = j;
                        while(a[y][i]){ //利用gcd的方法，不停地进行辗转相除，达到消去其他行对应列元素的目的
                            ll t = a[x][i] / a[y][i];
                            for(int k=i; k<=n; k++)
                                a[x][k] = (a[x][k] - a[y][k]*t)%mod;
                            swap(x,y);
                        }
                        
                        if(x != i){     //奇数次交换，则D=-D'整行交换
                            for(int k = 1; k<=n; k++){
                                swap(a[i][k], a[x][k]);
                            }
                            sign ^= 1;
                        }
                    }
                    if(a[i][i] == 0){   //斜对角中有一个0，则结果为0
                        puts("0");
                        return;
                    }
                    else ans = ans * a[i][i] %mod;
                }
                if(sign) ans *= -1;
                if(ans < 0) ans += mod;
                printf("%lld
", ans);
            }   
int main(){
            while(~scanf("%d%d", &n, &m)){
                k = 0;
                for(int i=1; i<=n; i++){
                        scanf("%s", str);
                        for(int j=0; j<m; j++){
                            if(str[j] == '.')g[i][j+1] = ++k;
                        }
                }

                for(int i=1; i<=n; i++){
                    for(int j=1; j<=m; j++){
                        int u = g[i][j],v;
                        if(u > 0){
                            if(i +1 <= n && g[i+1][j]){
                                v = g[i+1][j];
                                a[u][v] = a[v][u] = -1;
                                a[u][u]++;a[v][v]++;
                            }   
                            if(j + 1<=m && g[i][j+1]){
                                v = g[i][j+1];
                                a[u][v] = a[v][u] = -1;
                                a[u][u]++;a[v][v]++;
                            }
                        }
                    }
                }
                n = k-1;
                cal();
            }
            return 0;
}