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 = 505;
const int M = N * N;
const lf eps = 1e-7;
 
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 = 1; i <= n; ++i) {
        for (k = i, j = i + 1; 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 + 1; ++j) swap(a[i][j], a[k][j]);
        for (k = i + 1; k <= n; ++k)
            for (tmp = -a[k][i] / a[i][i], j = i; j <= n + 1; ++j)
                a[k][j] += a[i][j] * tmp;
    }
    for (i = n; i; --i) {
        for (j = i + 1; j <= n; ++j)
            a[i][n + 1] -= a[i][j] * ans[j];
        ans[i] = a[i][n + 1] / a[i][i];
    }
}
 
int main() {
    int i, j;
    n = read(), m = read();
    for (i = 1; i <= m; ++i)
        Add_Edges(read(), read());
    for (i = 1; i < n; ++i) {
        for (j = 1; j < n; ++j)
            if (deg[j]) a[i][j] = a[i][j] / deg[j];
        a[i][i] = 1, a[i][n] = 0;
    }
    a[1][n] = 1;
    gauss(n - 1);
    for (i = 1; i <= n; ++i)
        if (deg[i]) ans[i] /= deg[i];
    for (i = 1; i <= m; ++i)
        w[i] = ans[E[i].x] + ans[E[i].y];
    sort(w + 1, w + m + 1);
    for (i = 1; i <= m; ++i) Ans += w[i] * (m - i + 1);
    printf("%.3lf
", Ans);
    return 0;
}
 
inline int read() {
    static int x;
    static char ch;
    x = 0, ch = getchar();
    while (ch < '0' || '9' < ch)
        ch = getchar();
    while ('0' <= ch && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x;
}