P3317 [SDOI2014]重建 变元矩阵树定理 高斯消元

传送门：https://www.luogu.org/problemnew/show/P3317

这道题的推导公式还是比较好理解的，但是由于这个矩阵是小数的，要注意高斯消元方法的使用；

#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>

/*

⊂_ヽ
　 ＼＼ Λ＿Λ  来了老弟
　　 ＼(‘?‘)
　　　 >　⌒ヽ
　　　/ 　 へ＼
　　 /　　/　＼＼
　　 ?　ノ　　 ヽ_つ
　　/　/
　 /　/|
　(　(ヽ
　|　|、＼
　| 丿 ＼ ⌒)
　| |　　) /
‘ノ )　　L?

*/

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

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
//typedef __int128 bll;
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 ‘\n‘

#define boost ios::sync_with_stdio(false);cin.tie(0)
#define rep(a, b, c) for(int a = (b); a <= (c); ++ a)
#define max3(a,b,c) max(max(a,b), c);
#define min3(a,b,c) min(min(a,b), c);

const ll oo = 1ll<<17;
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;
}

inline void cmax(int &x,int y){if(x<y)x=y;}
inline void cmax(ll &x,ll y){if(x<y)x=y;}
inline void cmin(int &x,int y){if(x>y)x=y;}
inline void cmin(ll &x,ll y){if(x>y)x=y;}

/*-----------------------showtime----------------------*/
            const int maxn = 109;
            double mp[maxn][maxn];
            double a[maxn][maxn];

            int n;

            double Gauss(int n){
                for(int i=1; i<n; i++){
                    int mx = i;
                    for(int j=i+1; j<n; j++){
                        if(fabs(a[mx][i]) < fabs(a[j][i])) mx = j;
                    }
                    swap(a[i], a[mx]);
                    for(int j=i+1; j<n; j++){
                        double tmp = a[j][i]/a[i][i];
                        for(int k=i+1; k<n; k++)
                            a[j][k] -= a[i][k] * tmp;
                    }
                }
                double ans = 1;
                for(int i=1; i<n; i++) ans *= a[i][i];
                return fabs(ans);
            }

int main(){
            scanf("%d", &n);
            double tmp = 1;
            for(int i=1; i<=n; i++) {
                for(int j=1; j<=n; j++) {
                    scanf("%lf", &mp[i][j]);
                    if(fabs(1.0 - mp[i][j]) < esp) mp[i][j] = 1.0- esp;
                    if(i < j)tmp = tmp * (1.0-mp[i][j]);
                    mp[i][j] = mp[i][j] / (1.0 - mp[i][j]);
                }
            }

            for(int i=1; i<=n; i++){
                for(int j=i+1; j<=n; j++){
                    a[i][j] = a[j][i] = -mp[i][j];
                    a[i][i] += mp[i][j];
                    a[j][j] += mp[i][j];
                }
            }
            printf("%.10f\n", Gauss(n) * tmp);
            return 0;
}

原文地址：https://www.cnblogs.com/ckxkexing/p/10503509.html