「雜項」Nov. 26st, 2020 - Training REC
阿新 • • 發佈：2020-11-27
Topic：高斯消元
| Problems | States |
|---|---|
| #8799. 熄燈遊戲 | AC |
| #15778. Dotp 驅逐豬玀 | AC |
| #17270. 博物館 | AC |
| #16337. XOR 和路徑 | IP |
Problem. 1 Junior - Formula
用 \(m_{i,j}\) 表示答案，\(a_{i,j}\) 表示原矩陣。
則方程為 \((m_{i,j}+m_{i-1,j}+m_{i+1,j}+m_{i,j-1}+m_{i,j+1}+a_{i,j})\operatorname{bitand}1=0\)。
變下形高消即可。
#include <cstdio> #include <cstring> const int MAXN = 50; template<typename _T> void swapp ( _T &x, _T &y ) { _T w = x; x = y; y = w; } int mat[MAXN][MAXN], ori[MAXN][MAXN]; int wax[5] = { 0, 0, 0, 1, -1 }, way[5] = { 0, 1, -1, 0, 0 }; int has ( const int x, const int y ) { return ( x - 1 ) * 6 + y; } void Eradicate () { for ( int i = 1; i <= 30; ++ i ) { int p = i; for ( int j = i + 1; j <= 30; ++ j ) { if ( mat[p][i] < mat[j][i] ) p = j; } for ( int j = 1; j <= 31; ++ j ) swapp ( mat[p][j], mat[i][j] ); for ( int j = 1; j <= 30; ++ j ) { if ( i == j || ! mat[j][i] ) continue; for ( int k = 1; k <= 31; ++ k ) mat[j][k] ^= mat[i][k]; } } } int main () { int cases; scanf ( "%d", &cases ); for ( int _ = 1; _ <= cases; ++ _ ) { memset( mat, 0, sizeof ( mat ) ); for ( int i = 1; i <= 5; ++ i ) { for ( int j = 1; j <= 6; ++ j ) scanf ( "%d", &ori[i][j] ); } for ( int i = 1; i <= 5; ++ i ) { for ( int j = 1; j <= 6; ++ j ) { for ( int k = 0; k < 5; ++ k ) { int nxi = i + wax[k], nxj = j + way[k]; if ( nxi >= 1 && nxi <= 5 && nxj >= 1 && nxj <= 6 ) mat[has ( i, j )][has ( nxi, nxj )] = 1; } mat[has ( i, j )][31] = ori[i][j]; } } Eradicate (); printf ( "PUZZLE #%d\n", _ ); for ( int i = 1; i <= 5; ++ i ) { for ( int j = 1; j <= 6; ++ j ) printf ( "%d ", mat[has ( i, j )][31] ); putchar ( '\n' ); } } return 0; }
Problem. 2 Junior / Senior - Formula
直接設概率做不來的樣子。。。
問了一下懂行的人發現需要設期望。
設 \(f_{u}\) 為節點經過 \(u\) 的期望次數，設 \(E\) 為邊集，\(d_{u}\) 表示節點 \(u\) 的度。
\[f_{u}=(1-\frac{p}{q})\sum_{(u,v)\in E}\frac{1}{d_{v}}f_{v} \]因為有環，所以需要高斯消元一下。
那麼在每個節點爆炸的概率即為 \(f_{u}\times\frac{p}{q}\)。
總結一下，這道題告訴了我不僅是期望可以迴歸本質用概率搞，概率也可以用期望來整。
#include <cstdio> const int MAXN = 300 + 5, MAXM = MAXN * MAXN; template<typename _T> _T ABS ( const _T x ) { return x > 0 ? x : -x; } template<typename _T> void swapp ( _T &x, _T &y ) { _T w = x; x = y; y = x; } struct GraphSet { int to, nx; GraphSet ( int T = 0, int N = 0 ) { to = T, nx = N; } } as[MAXM]; int n, m, p, q, cnt, d[MAXN], begin[MAXN]; double mat[MAXN][MAXN], sol[MAXN], boomed, transed; void makeEdge ( const int u, const int v ) { as[++ cnt] = GraphSet ( v, begin[u] ), begin[u] = cnt; } void Eradicate () { for ( int i = 1; i <= n; ++ i ) { int p = i; for ( int j = i + 1; j <= n; ++ j ) { if ( ABS ( mat[p][i] ) < ABS ( mat[j][i] ) ) p = j; } for ( int j = 1; j <= n + 1; ++ j ) swapp ( mat[p][j], mat[i][j] ); for ( int j = 1; j <= n; ++ j ) { if ( i == j ) continue; double d = mat[j][i] / mat[i][i]; for ( int k = 1; k <= n + 1; ++ k ) mat[j][k] -= mat[i][k] * d; } } for ( int i = 1; i <= n; ++ i ) sol[i] = mat[i][n + 1] / mat[i][i]; } int main () { scanf ( "%d%d%d%d", &n, &m, &p, &q ); for ( int i = 1, u, v; i <= m; ++ i ) { scanf ( "%d%d", &u, &v ); makeEdge ( u, v ), makeEdge( v, u ); d[u] ++, d[v] ++; } boomed = ( double )p / ( double )q; transed = 1 - boomed; for ( int _ = 1; _ <= n; ++ _ ) { int u = _; mat[u][u] = 1; for ( int i = begin[u]; i; i = as[i].nx ) { int v = as[i].to; mat[u][v] -= transed / d[v]; } } mat[1][n + 1] = 1; Eradicate (); for ( int i = 1; i <= n; ++ i ) printf ( "%.9lf\n", sol[i] * boomed ); return 0; }
Problem. 3 Junior / Senior- Formula & Thinking
設 \(f_{u,v}\) 表示同一時刻 Petya 在點 \(u\)，Vasya 在點 \(v\) 的概率。
\[f_{u,v}=p_{u}p_{v}f_{u,v}+\sum_{(u,u')\in E}\sum_{(v,v')\in E}\frac{1-p_{u'}}{d_{u'}}\frac{1-p_{v'}}{d_{v'}}f_{u',v'}+\sum_{(u,u')\in E}p_{v}\frac{1-p_{u'}}{d_{v'}}f_{u',v}+\sum_{(v,v')\in E}p_{u}\frac{1-p_{v'}}{d_{v'}}f_{u,v'} \]完了。
#include <cstdio>
const double EPS = 1e-8;
const int MAXN = 500 + 5, MAXM = MAXN * MAXN;
template<typename _T> _T ABS ( const _T x ) { return x > 0 ? x : -x; }
template<typename _T> void swapp ( _T &x, _T &y ) { _T w = x; x = y; y = w; }
struct GraphSet {
int to, nx;
GraphSet ( int T = 0, int N = 0 ) { to = T, nx = N; }
} as[MAXM];
int n, m, s, t, cnt, upper, d[MAXN], begin[MAXN];
double mat[MAXN][MAXN], p[MAXN];
int has ( const int x, const int y ) { return ( x - 1 ) * n + y; }
void makeEdge ( const int u, const int v ) { as[++ cnt] = GraphSet ( v, begin[u] ), begin[u] = cnt; }
void Eradicate () {
for ( int i = 1; i <= upper; ++ i ) {
int p = i;
for ( int j = i + 1; j <= upper; ++ j ) {
if ( ABS ( mat[p][i] ) < ABS ( mat[j][i] ) ) p = i;
}
for ( int j = i + 1; j <= upper + 1; ++ j ) swapp ( mat[p][j], mat[i][j] );
if ( ABS ( mat[i][i] ) < EPS ) continue;
for ( int j = 1; j <= upper; ++ j ) {
if ( i == j ) continue;
double d = mat[j][i] / mat[i][i];
for ( int k = 1; k <= upper + 1; ++ k ) mat[j][k] -= mat[i][k] * d;
}
}
for ( int i = 1; i <= upper; ++ i ) mat[i][upper + 1] /= mat[i][i];
}
int main () {
scanf ( "%d%d%d%d", &n, &m, &s, &t ), upper = n * n;
for ( int i = 1, u, v; i <= m; ++ i ) {
scanf ( "%d%d", &u, &v );
makeEdge ( u, v ), makeEdge ( v, u );
d[u] ++, d[v] ++;
}
for ( int i = 1; i <= n; ++ i ) scanf ( "%lf", &p[i] );
for ( int i = 1; i <= n; ++ i ) {
for ( int j = 1; j <= n; ++ j ) {
if ( i == j ) mat[has ( i, j )][has ( i, j )] = 1;
else mat[has ( i, j )][has ( i, j )] = 1 - p[i] * p[j];
for ( int ii = begin[i]; ii; ii = as[ii].nx ) {
int u = as[ii].to;
for ( int iii = begin[j]; iii; iii = as[iii].nx ) {
int v = as[iii].to;
if ( u == v ) continue;
mat[has ( i, j )][has ( u, v )] = -( 1 - p[u] ) / d[u] * ( 1 - p[v] ) / d[v];
}
}
for ( int ii = begin[i]; ii; ii = as[ii].nx ) {
int u = as[ii].to;
if ( u == j ) continue;
mat[has ( i, j )][has ( u, j )] = -p[j] * ( 1 - p[u] ) / d[u];
}
for ( int ii = begin[j]; ii; ii = as[ii].nx ) {
int v = as[ii].to;
if ( i == v ) continue;
mat[has ( i, j )][has ( i, v )] = -p[i] * ( 1 - p[v] ) / d[v];
}
}
}
mat[has ( s, t )][upper + 1] = 1;
Eradicate ();
for ( int i = 1; i <= n; ++ i ) printf ( "%.9lf ", mat[has ( i, i )][upper + 1] );
return 0;
}