JZOJ 6385 B

考虑一般图上的随机游走问题，易列得方程 fu=1degu∑(u,v)∈Efv。
复杂度 O(n3)。

但是这一题运用了系数的一个特点：若从下往上（即从 n 到 1）消元，则消到 p 时，只有 p,fap,fafap 的方程中 fp 的系数不为 1。
故总共需要消元的方程是 O(n) 级别的。
复杂度 O(n2)。

代码：

#include <cstdio>
using namespace std;
const int N = 2e3;
const int mod = 998244353;
inline int fpow(int a,int b)
{
    int ret = 1;
    for(;b;b >>= 1)
        (b & 1) && (ret = (long long)ret * a % mod),a = (long long)a * a % mod;
    return ret;
}
int n,fa[N + 5],deg[N + 5];
int a[N + 5][N + 5],f[N + 5];
int to[(N << 2) + 5],pre[(N << 2) + 5],first[N + 5];
inline void add(int u,int v)
{
    static int tot = 0;
    to[++tot] = v,pre[tot] = first[u],first[u] = tot;
}
int main()
{
    freopen("b.in","r",stdin),freopen("b.out","w",stdout);
    scanf("%d",&n);
    for(register int i = 2;i <= n;++i)
        scanf("%d",fa + i),add(fa[i],i),add(i,fa[i]),++deg[fa[i]],++deg[i],fa[fa[i]] && (add(fa[fa[i]],i),add(i,fa[fa[i]]),++deg[fa[fa[i]]],++deg[i]);
    a[1][1] = 1;
    for(register int i = 2,inv;i <= n;++i)
    {
        inv = fpow(deg[i],mod - 2),a[i][0] = a[i][i] = 1;
        for(register int j = first[i];j;j = pre[j])
            a[i][to[j]] = mod - inv;
    }
    for(register int i = n,inv;i;--i)
    {
        if(fa[i])
        {
            inv = (long long)a[fa[i]][i] * fpow(a[i][i],mod - 2) % mod;
            for(register int j = 0;j <= n;++j)
                a[fa[i]][j] = (a[fa[i]][j] - (long long)a[i][j] * inv % mod + mod) % mod;
        }
        if(fa[fa[i]])
        {
            inv = (long long)a[fa[fa[i]]][i] * fpow(a[i][i],mod - 2) % mod;
            for(register int j = 0;j <= n;++j)
                a[fa[fa[i]]][j] = (a[fa[fa[i]]][j] - (long long)a[i][j] * inv % mod + mod) % mod;
        }
    }
    for(register int i = 1;i <= n;++i)
        for(register int j = first[i],inv;j;j = pre[j])
            if((to[j] ^ fa[i]) && (to[j] ^ fa[fa[i]]))
            {
                inv = (long long)a[to[j]][i] * fpow(a[i][i],mod - 2) % mod;
                for(register int k = 0;k <= n;++k)
                    a[to[j]][k] = (a[to[j]][k] - (long long)a[i][k] * inv % mod + mod) % mod;
            }
    for(register int i = 1;i <= n;++i)
        printf("%d\n",(long long)a[i][0] * fpow(a[i][i],mod - 2) % mod);
}