SP2885 WORDRING - Word Rings

本题求平均环长,可以将每个字符串的前后两个字符看做点,这两个点存在于一个字符串的关系视为一条边,接着二分枚举答案(单调性是很显然的),用队列优化BellmanFord判断即可.

关键代码:

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static bool __fastcall spfa_dfs (Re int fr, Re double mid)
{
vis[fr] = true;
register int to;
for (Re int i = head[fr]; i; i = nxt[i])
{
to = too[i];
if (dis[fr] + wei[i] - mid > dis[to])
{
dis[to] = dis[fr] + wei[i] - mid;
if (vis[to] || spfa_dfs(to, mid))
{vis[fr] = false; return true;}
}
}
vis[fr] = false; return false;
}
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inline bool judge (register double mid)
{
memset (dis, 0, sizeof dis);
for (Re int i = 1; i <= hhsh; ++ i)
if (spfa_dfs (i, mid)) return true;
return false;
}
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while (true)
{
n = read (); if (!n) return 0;
memset (head, 0, sizeof head);
memset (hash, 0, sizeof hash);
ecnt = hhsh = 0;
while (n --)
{
c = gc(), len = 0;
while (c == '\n' || c == '\r') c = gc ();
while ((c != '\n') && (c != '\r')) s[++ len] = c, c = gc();
fr = calc(s[1], s[2]), to = calc(s[len-1], s[len]);
if (!hash[fr]) hash[fr] = ++ hhsh;
if (!hash[to]) hash[to] = ++ hhsh;
addedge (hash[fr], hash[to], len);
}
l = 0.0, r = 1000.0;
while (r - l >= eps)
{
mid = (l + r) / 2;
if (judge(mid)) l = mid;
else r = mid;
}
if (l == 0) puts ("No solution.");
else printf ("%.2lf\n", l);
}

祝大家 RP++ !

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