BZOJ3143 [Hnoi2013]游走

首先高斯消元解出每个点被走到的概率

注意到这里走到$n$就停下来了，所以$P(n) = 0$

解出来以后，给每条边$(u, v)$赋边权$P(u) + P(v)$即可，然后直接贪心

 /**************************************************************
     Problem: 3143
     User: rausen
     Language: C++
     Result: Accepted
     Time:680 ms
     Memory:6792 kb
 ****************************************************************/

 #include <cstdio>
 #include <cmath>
 #include <algorithm>

 using namespace std;
 typedef double lf;
 const int N = ;
 const int M = N * N;
 const lf eps = 1e-;

 int read();

 int n, m, tot;
 int deg[N];
 lf a[N][N], ans[N], w[M];
 lf Ans;

 struct edge {
     int x, y;
 } E[M];

 inline void Add_Edges(int x, int y) {
     E[++tot].x = x, E[tot].y = y;
     a[x][y] = a[y][x] = -1.0;
     ++deg[x], ++deg[y];
 }

 void gauss(int n) {
     int i, j, k;
     lf tmp;
     for (i = ; i <= n; ++i) {
         for (k = i, j = i + ; j <= n; ++j)
             if (fabs(a[j][i]) > fabs(a[k][i])) k = j;
         if (fabs(a[k][i]) < eps) continue;
         for (j = i; j <= n + ; ++j) swap(a[i][j], a[k][j]);
         for (k = i + ; k <= n; ++k)
             for (tmp = -a[k][i] / a[i][i], j = i; j <= n + ; ++j)
                 a[k][j] += a[i][j] * tmp;
     }
     for (i = n; i; --i) {
         for (j = i + ; j <= n; ++j)
             a[i][n + ] -= a[i][j] * ans[j];
         ans[i] = a[i][n + ] / a[i][i];
     }
 }

 int main() {
     int i, j;
     n = read(), m = read();
     for (i = ; i <= m; ++i)
         Add_Edges(read(), read());
     for (i = ; i < n; ++i) {
         for (j = ; j < n; ++j)
             if (deg[j]) a[i][j] = a[i][j] / deg[j];
         a[i][i] = , a[i][n] = ;
     }
     a[][n] = ;
     gauss(n - );
     for (i = ; i <= n; ++i)
         if (deg[i]) ans[i] /= deg[i];
     for (i = ; i <= m; ++i)
         w[i] = ans[E[i].x] + ans[E[i].y];
     sort(w + , w + m + );
     for (i = ; i <= m; ++i) Ans += w[i] * (m - i + );
     printf("%.3lf\n", Ans);
     return ;
 }

 inline int read() {
     static int x;
     static char ch;
     x = , ch = getchar();
     while (ch < '' || '' < ch)
         ch = getchar();
     while ('' <= ch && ch <= '') {
         x = x *  + ch - '';
         ch = getchar();
     }
     return x;
 }