多类型高斯消元杂题

[SDOI2010]外星千足虫
- description
- solution
- code
[HNOI2013]游走
- description
- solution
- code
[HNOI2011]XOR和路径
- description
- solution
- code
Maze(树上高斯消元)
- problem
- solution
- code

[SDOI2010]外星千足虫

solution

高斯消元的模板题

虽然感觉问了个最早确定所有虫的时间戳，但并没有什么用

在高斯消元过程中，使用的最大行编号就是最早确定时间

code

#include <cstdio>
#include <bitset>
#include <iostream>
using namespace std;
#define maxn 1005
int n, m;
bitset < maxn > a[maxn << 1];int main() {scanf( "%d %d", &n, &m );for( int i = 1, x;i <= m;i ++ ) {for( int j = 1;j <= n;j ++ )scanf( "%1d", &x ), a[i][j] = x;scanf( "%d", &x ), a[i][n + 1] = x;}int ans = 0;for( int i = 1;i <= n;i ++ ) {int row = i;while( row <= m && ! a[row][i] ) row ++;if( row == m + 1 ) return ! printf( "Cannot Determine\n" );if( i ^ row ) swap( a[row], a[i] );ans = max( ans, row );for( int j = 1;j <= m;j ++ )if( i == j || ! a[j][i] ) continue;else a[j] ^= a[i];}printf( "%d\n", ans );for( int i = 1;i <= n;i ++ )if( a[i][n + 1] ) printf( "?y7M#\n" );else printf( "Earth\n" );return 0;
}

[HNOI2013]游走

solution

将路径拆分成每条边期望经过次数乘以边权的求和

贪心的把最大权值放在期望经过次数最少的边上

但是边的级别是空间时间不足以承受的

事实上，边的期望至于边连接的两个端点有关

每个点到与之相连边的概率一样，期望一样

所以到某条特定边的期望就是经过该点期望除以该点连接的总边数

一条边被经过的期望则是两个端点到这条边的期望和

于是乎求边的期望就转化为求点的期望

而点的期望至于相邻点有关

设 $E_i:i$ 点期望， $num_i:i$ 点总边数， $x_1,x_2,...,x_k$ 与 $i$ 相连，则有 $Ei=∑j=1kEjnumjE_i=\sum_{j=1}^k\frac{E_j}{num_j}$

每个点都能列出一个这样的方程，高斯消元解决

特别地，游走是从点 $1$ 开始的，所以期望要 $+ 1$ ，游走到点 $n$ 就不会继续进行了，所以计算期望时不能纳入考虑

code

#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
#define maxn 505
#define eps 1e-6
#define maxm 125005
vector < int > G[maxn]; 
int n, m;
int num[maxn], from[maxm], to[maxm];
double E[maxm];
double x[maxn][maxn];double Fabs( double x ) {return x < 0 ? -x : x;
}int main() {scanf( "%d %d", &n, &m );for( int i = 1, u, v;i <= m;i ++ ) {scanf( "%d %d", &u, &v );G[u].push_back( v );G[v].push_back( u );from[i] = u, to[i] = v;num[u] ++, num[v] ++;}x[1][n] = 1;for( int i = 1;i < n;i ++ ) {x[i][i] = 1;for( auto j : G[i] )if( j != n )x[i][j] = -1.0 / num[j];}for( int i = 1;i < n;i ++ ) {int k = i;for( int j = i + 1;j < n;j ++ )if( Fabs( x[j][i] ) > Fabs( x[k][i] ) )k = j;if( i ^ k ) swap( x[i], x[k] );for( int j = n;j >= i;j -- )x[i][j] /= x[i][i];for( int j = 1;j < n;j ++ )if( i ^ j )for( int k = n;k >= i;k -- )x[j][k] -= x[j][i] * x[i][k];}for( int i = 1;i <= m;i ++ ) {if( from[i] != n )E[i] += x[from[i]][n] / num[from[i]];if( to[i] != n )E[i] += x[to[i]][n] / num[to[i]];}sort( E + 1, E + m + 1 );double ans = 0;for( int i = 1;i <= m;i ++ )ans += E[i] * ( m - i + 1 );printf( "%.3f", ans );return 0;
}

[HNOI2011]XOR和路径

solution

一般看到充满二进制意味的异或都要考虑拆位

此题也不例外，每一个单独计算期望

设 $f_{i}:$ 表示 $i→ni\rightarrow n$ 路径该位为 $1$ 的概率，则 $1-f_i$ 表示 $i→ni\rightarrow n$ 路径为 $0$ 的概率， $d_i:i$ 的出度

$∀(u,v)∈Edgefu=1du(∑w(u,v)=0fv+∑w(u,v)=1(1−fv))⇔∀(u,v)∈Edgedu×fu=(∑w(u,v)=0fv+∑w(u,v)=1(1−fv))\forall_{(u,v)\in Edge}f_u=\frac{1}{d_u}\Big(\sum_{w(u,v)=0}f_v+\sum_{w(u,v)=1}(1-f_v)\Big)\Leftrightarrow \forall_{(u,v)\in Edge}d_u\times f_u=\Big(\sum_{w(u,v)=0}f_v+\sum_{w(u,v)=1}(1-f_v)\Big)$
在这里插入图片描述

换言之， $u→nu\rightarrow n$ 的路径想要为 $1 :$ 如果 $(u→v)(u\rightarrow v)$ 这条边为 $1$ ，必须 $v→nv\rightarrow n$ 的路径为 $0$ ，否则为 $1$

方程化为

$dufu−∑w(u,v)=0fv+∑w(u,v)=1fu=∑w(u,v)=11d_uf_u-\sum_{w(u,v)=0}f_v+\sum_{w(u,v)=1}f_u=\sum_{w(u,v)=1}1$

高斯消元， $ans=∑i2ifi(1)ans=\sum_i2^if_i(1)$

code

#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
using namespace std;
#define eps 1e-8
#define maxn 105
vector < pair < int, int > > G[maxn];
int n, m;
int d[maxn];
double ans[maxn];
double a[maxn][maxn];double Fabs( double x ) {return x < 0 ? -x : x;
}void build( int x ) {a[n][n] = 1;for( int i = 1;i < n;i ++ ) {a[i][i] = d[i];for( auto j : G[i] ) {if( j.second & x ) a[i][j.first] ++, a[i][n + 1] ++;else a[i][j.first] --;		}}
}void gauss() {for( int i = 1;i <= n;i ++ ) {int k = i;for( int j = i;j <= n;j ++ )if( Fabs( a[j][i] ) > Fabs( a[k][i] ) )k = j;if( k ^ i ) swap( a[i], a[k] );for( int j = i + 1;j <= n;j ++ )if( Fabs( a[j][i] ) < eps ) continue;else {double t = a[j][i] / a[i][i];for( k = i;k <= n + 1;k ++ )a[j][k] -= a[i][k] * t;a[j][i] = 0;}}for( int i = n;i;i -- ) {for( int j = i + 1;j <= n;j ++ )a[i][n + 1] -= a[i][j] * ans[j];ans[i] = a[i][n + 1] / a[i][i];}memset( a, 0, sizeof( a ) );
}int main() {scanf( "%d %d", &n, &m );int maxx = 0;for( int i = 1, u, v, w;i <= m;i ++ ) {scanf( "%d %d %d", &u, &v, &w );G[u].push_back( make_pair( v, w ) );d[u] ++;if( u ^ v ) {G[v].push_back( make_pair( u, w ) );d[v] ++;}maxx = max( maxx, w );}double ret = 0;for( int i = 1;i <= maxx;i <<= 1 ) {build( i );gauss();ret += ans[1] * i;}printf( "%.3f\n", ret );return 0;
}

Maze(树上高斯消元)

problem

solution

令 $E_i:i$ 节点逃出期望经过的边数， $edge_i:i$ 相连边数
$∀i,i∈leafEi=ki×E1+ei×0+(1−ki−ei)×(Efai+1)=ki×E1+(1−ki−ei)×Efai+(1−ki−ei)\forall_{i,i\in leaf}E_i=k_i\times E_1+e_i\times 0+(1-k_i-e_i)\times (E_{fa_i}+1)\\=k_i\times E_1+(1-k_i-e_i)\times E_{fa_i}+(1-k_i-e_i)$

$∀i,i∉leafEi=ki×E1+ei×0+(1−ki−ei)×1edgei×(∑j∈soni(Ej+1)+Efai+1)=ki×E1+1−ki−eiedgei×Efai+1−ki−eim×∑j∈soniEj+(1−ki−ei)\forall_{i,i\notin leaf}E_{i}=k_i\times E_1+e_i\times 0+(1-k_i-e_i)\times\frac{1}{edge_i}\times\Big(\sum_{j\in son_i}(E_j+1)+E_{fa_i}+1\Big)\\=k_i\times E_1+\frac{1-k_i-e_i}{edge_i}\times E_{fa_i}+\frac{1-k_i-e_i}{m}\times\sum_{j\in son_i}E_j+(1-k_i-e_i)$

转移涉及父亲儿子且彼此依赖，而我们非常想得到不交叉的线性递推关系

想办法变成只跟父亲有关的转移，形式的设 $Ei=Ai×E1+Bi×Efai+CiE_i=A_i\times E_1+B_i\times E_{fa_i}+C_i$

类比解方程组，列出以下式子

$∀i,i∉leaf\forall_{i,i\notin leaf}$
$∑j∈soniEj=∑j∈soniAj×E1+Bj×Efaj+Cj=∑j∈soniAj×E1+Bj×Ei+Cj\sum_{j\in son_i}E_j=\sum_{j\in son_i}A_j\times E_1+B_j\times E_{fa_j}+C_j=\sum_{j\in son_i}A_j\times E_1+B_j\times E_i+C_j\\$

$Ei=kiE1+1−ki−eiedgeiEfai+1−ki−eim∑j∈soniEj+(1−ki−ei)=kiE1+1−ki−eiedgeiEfai+1−ki−eiedgei∑j∈soni(AjE1+BjEi+Cj)+(1−ki−ei)E_i=k_iE_1+\frac{1-k_i-e_i}{edge_i}E_{fa_i}+\frac{1-k_i-e_i}{m}\sum_{j\in son_i}E_j+(1-k_i-e_i)\\=k_iE_1+\frac{1-k_i-e_i}{edge_i}E_{fa_i}+\frac{1-k_i-e_i}{edge_i}\sum_{j\in son_i}(A_jE_1+B_jE_i+C_j)+(1-k_i-e_i)$
在这里插入图片描述

$⇔\Leftrightarrow$
$(1−1−ki−eiedgei∑j∈soniBj)Ei=(1-\frac{1-k_i-e_i}{edge_i}\sum_{j\in son_i}B_j)E_i=$ $(ki+1−ki−eiedgei∑j∈soniAj)E1+1−ki−eiedgeiEfai+(1−ki−ei)+1−ki−eiedgei∑j∈soniCj(k_i+\frac{1-k_i-e_i}{edge_i}\sum_{j\in son_i}A_j)E_1+\frac{1-k_i-e_i}{edge_i}E_{fa_i}+(1-k_i-e_i)+\frac{1-k_i-e_i}{edge_i}\sum_{j\in son_i}C_j$
$⇒Ai=ki+1−ki−eiedgei∑j∈soniAj(1−1−ki−eiedgei∑j∈soniBj)\Rightarrow A_i=\frac{k_i+\frac{1-k_i-e_i}{edge_i}\sum_{j\in son_i}A_j}{(1-\frac{1-k_i-e_i}{edge_i}\sum_{j\in son_i}B_j)}$ $Bi=1−ki−eiedgei(1−1−ki−eiedgei∑j∈soniBj)B_i=\frac{\frac{1-k_i-e_i}{edge_i}}{(1-\frac{1-k_i-e_i}{edge_i}\sum_{j\in son_i}B_j)}$ $Ci=(1−ki−ei)+1−ki−eiedgei∑j∈soniCj(1−1−ki−eiedgei∑j∈soniBj)C_i=\frac{(1-k_i-e_i)+\frac{1-k_i-e_i}{edge_i}\sum_{j\in son_i}C_j}{(1-\frac{1-k_i-e_i}{edge_i}\sum_{j\in son_i}B_j)}$
在这里插入图片描述

$∀i,i∈leaf\forall_{i,i\in leaf}$
$Ei=ki×E1+(1−ki−ei)×Efai+(1−ki−ei)=Ai×E1+Bi×Efai+Ci⇒{Ai=kiBi=1−ki−eiCi=1−ki−eiE_i=k_i\times E_1+(1-k_i-e_i)\times E_{fa_i}+(1-k_i-e_i)=A_i\times E_1+B_i\times E_{fa_i}+C_i\\ \Rightarrow \begin{cases}A_i=k_i\\B_i=1-k_i-e_i\\C_i=1-k_i-e_i\end{cases}$
从叶子节点开始倒着往上递推（也被称之为树上高斯消元）

树上高斯消元

把有关父亲和儿子的递推式通过 $A_ix+B_iy+C_i$ 的形式转化为只由一方单向线性递推

然后正着/倒着递推消元

直到算出 $A_1,B_1,C_1$

$E1=A1×E1+B1×0+C1⇒E1=C11−A1E_1=A_1\times E_1+B_1\times 0+C_1\Rightarrow E_1=\frac{C_1}{1-A_1}$

当 $A_1$ 无限趋近于 $1$ 时，无解

如果 $i$ 的 $A_i,B_i,C_i$ 的分母形式 $1−1−ki−eiedgei∑j∈soniBj1-\frac{1-k_i-e_i}{edge_i}\sum_{j\in son_i}B_j$ 趋近于 $0$ ，也是无解

注意精度问题，eps=1e-8都不行，因为涉及 $/ 100$

code

#include <cmath>
#include <cstdio>
#include <vector>
using namespace std;
#define eps 1e-10
#define maxn 10005
vector < int > G[maxn];
int T, n;
double k[maxn], e[maxn], A[maxn], B[maxn], C[maxn];bool dfs( int i, int fa ) {if( G[i].size() == 1 && G[i][0] == fa ) {A[i] = k[i], B[i] = C[i] = 1 - k[i] - e[i];return 1;}A[i] = k[i];B[i] = ( 1 - k[i] - e[i] ) / G[i].size();C[i] = 1 - k[i] - e[i];double t = 0;for( auto j : G[i] ) {if( j == fa ) continue;else if( ! dfs( j, i ) ) return 0;A[i] += A[j] * B[i];C[i] += C[j] * B[i];t += B[j] * B[i];}if( fabs( 1 - t ) < eps ) return 0;else {A[i] /= ( 1 - t );B[i] /= ( 1 - t );C[i] /= ( 1 - t );return 1;}
}int main() {scanf( "%d", &T );for( int Case = 1;Case <= T;Case ++ ) {scanf( "%d", &n );for( int i = 1;i <= n;i ++ )G[i].clear();for( int i = 1, u, v;i < n;i ++ ) {scanf( "%d %d", &u, &v );G[u].push_back( v );G[v].push_back( u );  }for( int i = 1;i <= n;i ++ ) {scanf( "%lf %lf", &k[i], &e[i] );k[i] /= 100, e[i] /= 100;}if( dfs( 1, 0 ) && fabs( 1 - A[1] ) > eps )printf( "Case %d: %f\n", Case, C[1] / ( 1 - A[1] ) );elseprintf( "Case %d: impossible\n", Case );}return 0;
}