bzoj 3143 随机游走

题意:

　　给一个简单无向图,一个人从1号节点开始随机游走(即以相同概率走向与它相邻的点),走到n便停止,问每条边期望走的步数.

首先求出每个点期望走到的次数,每条边自然是从它的两个端点走来.

 /**************************************************************
     Problem: 3143
     User: idy002
     Language: C++
     Result: Accepted
     Time:736 ms
     Memory:9956 kb
 ****************************************************************/

 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 #define N 510
 #define M N*N
 using namespace std;

 int n, m;
 int head[N], dest[M], next[M], etot;
 int dgr[N], uu[M], vv[M], qu[M];
 double ww[M];
 double a[N][N];

 void adde( int u, int v ) {
     etot++;
     dest[etot] = v;
     next[etot] = head[u];
     head[u] = etot;
 }
 void gauss() {
     int i, j, k;
     for( i=; i<=n; i++ ) {
         j=i;
         for( k=i+; k<=n; k++ )
             if( fabs(a[k][i])>fabs(a[j][i]) ) j=k;
         for( k=i; k<=n+; k++ )
             swap( a[j][k], a[i][k] );
         for( j=i+; j<=n; j++ ) {
             double r = a[j][i]/a[i][i];
             for( k=i; k<=n+; k++ )
                 a[j][k] -= a[i][k]*r;
         }
     }
     for( int i=n; i>=; i-- ) {
         a[i][n+] /= a[i][i];
         a[i][i] = 1.0;
         for( int j=i-; j>=; j-- ) {
             a[j][n+] -= a[j][i]*a[i][n+];
             a[j][i] = 0.0;
         }
     }
 }
 bool cmp( int a, int b ) {
     return ww[a] > ww[b];
 }
 int main() {
     scanf( "%d%d", &n, &m );
     for( int i=; i<=m; i++ ) {
         scanf( "%d%d", uu+i, vv+i );
         adde( uu[i], vv[i] );
         adde( vv[i], uu[i] );
         dgr[uu[i]]++;
         dgr[vv[i]]++;
     }
     for( int i=; i<=n; i++ )
         a[i][i] = -1.0;
     a[][n+] = -;
     for( int u=; u<=n; u++ ) {
         for( int t=head[u]; t; t=next[t] ) {
             int v=dest[t];
             if( v==n ) continue;
             a[u][v] += 1.0/dgr[v];
         }
     }
     gauss();
     for( int i=; i<=m; i++ ) {
         int u=uu[i], v=vv[i];
         if( u!=n ) ww[i]+=a[u][n+]/dgr[u];
         if( v!=n ) ww[i]+=a[v][n+]/dgr[v];
     }
     for( int i=; i<=m; i++ )
         qu[i] = i;
     sort( qu+, qu++m, cmp );
     double ans = 0.0;
     for( int i=; i<=m; i++ )
         ans += i * ww[qu[i]];
     printf( "%.3lf\n", ans );
 }