A

拓撲排序+倍增哈希
或者
拓撲排序對於每個點計一個rank，每個點優先選取rank靠前的最小邊權點
每次依然按照rank排序更新rank

#include<bits/stdc++.h>
using namespace std;
 
template <typename T> void chmax(T &x,const T &y)
{
    if(x<y)x=y;
}
template <typename T> void chmin(T &x,const T &y)
{
    if(x>y)x=y;
}
typedef long long s64;
typedef unsigned long long u64;
typedef pair<int,int> pii;
 
#define rep(i,l,r) for(int i=l;i<=r;++i)
#define per(i,r,l) for(int i=r;i>=l;--i)
 
int rand_32()
{
    return (RAND_MAX<=32768)?(rand()*32768+rand()):rand();
}
u64 rand_64()
{
    return ((u64)rand_32()<<30)+rand_32();
}
 
const int N=1e6+5,L=20,D=998244353;
vector<pii>lk[N],nlk[N];
int to_fa[N];
int fa[N][L];u64 h[N][L];
int f[N];s64 g[N];
u64 w1[L],w2[L];
 
int du[N],q[N];
 
int smaller(int x,int y)
{
    per(j,L-1,0)
    if(h[x][j]==h[y][j])
    {
        x=fa[x][j];y=fa[y][j];
    }
    return to_fa[x]<to_fa[y];
}
 
int main()
{
//  freopen("1.in","r",stdin);freopen("std.out","w",stdout);
    int n,m;
    cin>>n>>m;
    rep(i,0,L-1){w1[i]=rand_64();w2[i]=rand_64();}
    rep(i,1,m)
    {
        int x,y,w;
        scanf("%d%d%d",&x,&y,&w);
        lk[x].push_back(pii(y,w));
        nlk[y].push_back(pii(x,w));
        ++du[x];
    }
    //cerr<<clock()<<endl;
    int tail=0;
    rep(x,1,n)
    if(!du[x])q[++tail]=x;
    rep(head,1,tail)
    {
        int x=q[head];
        for(auto pr:nlk[x])
        {
            int y=pr.first;
            if(!--du[y])q[++tail]=y;
        }
        if(lk[x].empty())continue;
        for(auto pr:lk[x])
        {
            int y=pr.first,w=pr.second;
            chmax(f[x],f[y]+1);
        }
        to_fa[x]=1e9;
        for(auto pr:lk[x])
        {
            int y=pr.first,w=pr.second;
            if(f[x]==f[y]+1)chmin(to_fa[x],w);
        }
        for(auto pr:lk[x])
        {
            int y=pr.first,w=pr.second;
            if(f[x]==f[y]+1&&to_fa[x]==w)
            if(!fa[x][0]||smaller(y,fa[x][0]))fa[x][0]=y;
        }
        g[x]=29*(g[fa[x][0]]+to_fa[x])%D;
        rep(j,1,L-1)fa[x][j]=fa[fa[x][j-1]][j-1];
        h[x][0]=to_fa[x];
        rep(j,1,L-1)h[x][j]=(h[x][j-1]^w2[j])*w1[j]+h[fa[x][j-1]][j-1];
    }  
    //cerr<<clock()<<endl;
    //cerr<<cnt<<endl;
    rep(x,1,n)
    if(du[x])puts("Infinity");
    else printf("%d\n",int(g[x]));
}
 

B
概率dp
對於個人的每個mx的每個選項考慮

#include<map>
#include<queue> 
#include<cstdio> 
#include<cstring> 
#include<algorithm> 
using namespace std; 
#define rep(a,b,c) for(int a = b; a <= c;++ a) 
#define per(a,b,c) for(int a = b; a >= c; -- a) 
#define gc getchar()
#define pc putchar
inline int read() { 
    int x = 0,f = 1; 
    char c = getchar(); 
    while(c < '0' || c > '9') c = getchar(); 
    while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = getchar(); 
    return x * f; 
} 
void print(int x) { 
    if(x < 0) { 
        putchar('-'); 
        x = -x; 
    } 
    if(x >= 10) print(x / 10); 
    putchar(x % 10 + '0'); 
} 
#define LL long long 
const int mod = 998244353; 
const int maxn = 2007; 
int n; 
int P[maxn][4],W[4][4]; 
LL ans[maxn]; 
LL f[maxn],g[maxn]; // all / base x
inline void m(LL &x) { 
    x = x >= mod ? x - mod : x; 
} 
inline int fstpow(int x,int k) { 
    int ret = 1; 
    for(;k;k >>= 1,x = 1ll * x * x % mod)   
        if(k & 1) ret = 1ll * ret * x % mod; 
    return ret; 
} 
void solve(int mx) { 
    memset(f,0,sizeof f); 
    f[0] = 1; 
    rep(i,1,n) { 
        LL p = P[i][mx]; 
        per(j,i,1) 
            f[j] = (f[j - 1] * p % mod + f[j] * (1 + mod - p) % mod) % mod; 
        f[0] = f[0] * (1 + mod - p) % mod;  
    } 
    rep(i,1,n) { 
        LL p = P[i][mx]; 
        if(p != 1) { 
            LL inv = fstpow(mod + 1 - p,mod - 2); 
            g[0] = 1ll * f[0] * inv % mod; 
            rep(j,1,n) g[j] = 1ll * (f[j] - 1ll * p * g[j - 1] % mod) * inv % mod; 
        } 
        else 
            {rep(j,0,n - 1) g[j] = f[j + 1];g[n] = 0; }
        LL sum = 0; 
        rep(j,n / 2 + 1,n) sum += g[j]; 
        sum %= mod;  
        rep(j,0,3) (ans[i] += 1ll * W[mx][j] * P[i][j] % mod * (sum + (mx == j ? g[n / 2] : 0)) % mod) %= mod; 
    } 
} 
int main() { 
    n = read(); 
    rep(i,1,n) 
        P[i][0] = read(),P[i][1] = read(),P[i][2] = read(),P[i][3] = read(); 
    rep(i,0,3)
        W[i][0] = read(),W[i][1] = read(),W[i][2] = read(),W[i][3] = read(); 
    rep(i,0,3) solve(i); 
    for(int i = 1;i <= n;++ i) print((ans[i] % mod + mod) % mod),pc('\n'); 
    return 0; 
} 
 

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