01分数问题

#include <bits/stdc++.h>
using namespace std;
const int N = 700, M = 1e5 + 10;
int h[N], e[M], ne[M], w[M], idx;
int n;
double dist[N];
int cnt[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 check(double mid)
{
    queue<int> q;
    memset(dist, 0, sizeof dist);
    memset(st, 0, sizeof st);
    memset(cnt, 0, sizeof cnt);

    for (int i = 0; i < 676; i ++)
    {
        q.push(i);
        st[i] = true;
    }
    int count = 0;//用来结束spfa算法,如果count的计数大于某个数，我们判断图中一定有环
    while (q.size())
    {
        int t = q.front();
        q.pop();
        st[t] = false;
        for (int i = h[t]; ~i; i = ne[i])
        {
            int j = e[i];
            if (dist[j] < dist[t] + w[i] - mid)
            {
                dist[j] = dist[t] + w[i] - mid;
                cnt[j] = cnt[t] + 1;
                if (++ count > 10000) return true;
                if (cnt[j] >= N) return true;
                if (!st[j])
                {
                    q.push(j);
                    st[j] = true;
                }
            }
        }
    }
    return false;
}

int main(void)
{
    while (scanf("%d", &n), n)
    {
        memset(h, -1, sizeof h);
        idx = 0;
        char s[1010];
        while (n --)
        {
            scanf("%s", s);
            int len = strlen(s);
            if (len >= 2)
            {
                int a = (s[0] - 'a') * 26 + s[1] - 'a';
                int b = (s[len - 2] - 'a') * 26 + s[len - 1] - 'a';
                add(a, b, len);
            }
        }

        if (!check(0)) puts("No solution");
        else
        {
            double l = 0, r = 1000;
            while (r - l > 1e-4)
            {
                double mid = (l + r) / 2;
                if (check(mid)) l = mid;
                else r = mid;
            }
            printf("%lf\n", l);
        }
    }

    return 0;
}