(3ms)
将x23看作值为0的源点建图，在遍历到边的时候根据当前枚举的x23的值临时计算边的权

C++ 代码

#include<iostream>
#include<cstring>

using namespace std;

const int N = 26, M = 100;
int r[24], c[24], n, m;
int idx, h[N], w[M], e[M], ne[M], d[N], cnt[N], q[N];
bool st[N];
void add(int a, int b, int c)
{
    e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
}

bool spfa()
{
    memset(cnt, 0, sizeof cnt);
    memset(d, -0x3f, sizeof d);
    memset(st, 0, sizeof st);
    int hh = 0, tt = 1;
    q[0] = 23;
    d[23] = 0;
    while(hh != tt)
    {
        int u = q[hh++];
        if(hh == N) hh = 0;
        st[u] = false;
        for(int i = h[u]; ~i; i = ne[i])
        {
            int v = e[i], w0 = w[i];
            if(u == 23 && v == 22) w0 = w[i] + m;
            else if((v == 23 && (u == 22 || u == 15)) || (v < 7 && u == v + 16)) w0 = w[i] - m;
            else w0 = w[i];
            if(d[v] < d[u] + w0)
            {
                d[v] = d[u] + w0;
                cnt[v] = cnt[u] + 1;
                if(cnt[v] >= 24) 
                    return true;
                if(!st[v])
                {
                    st[v] = true;
                    q[tt++] = v;
                    if(tt == N) tt = 0;
                }
            }
        }
    }
    return false;
}

int main()
{
    int T;
    cin >> T;
    while(T--)
    {
        for(int i = 0; i < 24; ++i) cin >> r[i];
        cin >> n;
        for(int i = 0, t; i < n; ++i)
        {
            cin >> t;
            ++c[t];
        }
        idx = 0;
        memset(h, -1, sizeof h);
        for(int i = 0; i < 23; ++i)
        {
            add(i, i + 1, 0);
            add(i + 1, i, -c[i + 1]);
        }

        for(int i = 0; i < 24; ++i)
            add((i + 16) % 24, i, r[i]);

        add(23, 0, 0), add(0, 23, -c[0]);
        int lo = 1, hi = n + 1;
        while(lo < hi)
        {
            m = lo + hi >> 1;
            if(spfa()) lo = m + 1;
            else hi = m;
        }
        if(lo == n + 1) puts("No Solution");
        else printf("%d\n", lo);
    }
}