Bzoj 3270 Museum expected dp+ Gaussian elimination element

Topic: Given a non-connected graph, two people initially at a certain point, each moment each person will not move or choose to walk, ask two people finally where to meet

The two points (x, y) on the original map become a point on the new graph.

Then make a for the adjacency matrix of the graph (if two people on the same point there is no edge, otherwise by probability transfer), S is the initial row vector (s[(A, b)]=1), ans is the answer line vector

Then there are ans=s+sa+sa^2+sa^3+ ...

=s (i-a^+∞)/(I-A)

=s/(I-A)

So there ans* (i-a) =s

So the i-a of the transfer of Gaussian elimination element can be.

It's like the idea of expelling a pig.

#include <cmath> #include <cstdio> #include <cstring> #include <iomanip> #include <iostream > #include <algorithm> #define M 440#define P (x, y) (*n-n+ (y)) using namespace Std;struct abcd{int to,next;} Table[m];int head[m],tot;int n,m,a,b;int Degree[m];long double p[m],f[m][m],ans[m];void Add (int x,int y) {Table[++tot]. To=y;table[tot].next=head[x];head[x]=tot;} void gauss_elimination (int n) {int i,j,k;for (i=1;i<=n;i++) {k=i;for (j=i;j<=n;j++) if (Fabs (F[j][i]) >fabs (f[k ][i]) k=j;for (j=i;j<=n+1;j++) swap (f[i][j],f[k][j]); for (k=i+1;k<=n;k++) {long double temp=-f[k][i]/f[i][i]; for (j=i;j<=n+1;j++) f[k][j]+=f[i][j]*temp;}} for (i=n;i;i--) {for (j=i+1;j<=n;j++) f[i][n+1]-=f[i][j]*ans[j];ans[i]=f[i][n+1]/f[i][i];}} BOOL Compare (int x,int y) {return ans[p (x,x)] < Ans[p (y,y)];} int main () {//freopen ("electric.in", "R", stdin);//freopen ("Electric.out", "w", stdout); int i,j,x,y;cin>>n> >m>>a>>b;for (i=1;i<=m;i++) {scanf ("%d%d", &x,&y);d egree[x]++;d egree[y]++; ADD (x, y); ADD (y,x);} for (i=1;i<=n;i++) cin>>p[i];for (x=1;x<=n;x++) for (y=1;y<=n;y++) if (x!=y) {f[p (x, y)][p (x, y)]-=p[x]*p [Y];for (I=head[x];i;i=table[i].next) f[p (x, y)][p (table[i].to,y)]-= (1-p[x])/degree[x]*p[y];for (j=head[y];j;j= Table[j].next) f[p (x, y)][p (x,table[j].to)]-= (1-p[y])/degree[y]*p[x];for (I=head[x];i;i=table[i].next) for (j=head[ Y];j;j=table[j].next) f[p (x, y)][p (table[i].to,table[j].to)]-= (1-p[x]) * (1-p[y])/degree[x]/degree[y];} for (i=1;i<=n*n;i++) for (j=1;j<i;j++) swap (F[i][j],f[j][i]), for (i=1;i<=n*n;i++) f[i][i]+=1;f[p (A, b)][n*n+ 1]=1; Gauss_elimination (N*n); for (i=1;i<=n;i++) cout<<fixed<<setprecision (6) <<ans[p (i,i)]<< "; return 0;}

Bzoj 3270 Museum expected dp+ Gaussian elimination element