Codeforces 147 B. Smile House

题目链接：http://codeforces.com/contest/147/problem/B

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求有向图的最小正权环的大小   ${n<=300}$

非常显然的有${n^{3}log^2}$的做法，令${f_{(s,i,j)}}$表示从第${i}$个点走到第${j}$个点走了${2^{s}}$步的最大权值，然后二分答案，每次利用$f$数组计算答案。

问题在于算一下时间感觉是不太对的...然而居然跑过去了 QwQ

 #include<iostream>
 #include<cstdio>
 #include<algorithm>
 #include<vector>
 #include<cstdlib>
 #include<cmath>
 #include<cstring>
 using namespace std;
 #define maxn 310
 #define llg int
 #define inf 0x3f3f3f3f
 #define yyj(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);
 llg n,m,a[][maxn][maxn],dp[][maxn][maxn],up;

 inline int getint()
 {
     int w=,q=; char c=getchar();
     while((c<'' || c>'') && c!='-') c=getchar(); if(c=='-') q=,c=getchar();
     while (c>='' && c<='') w=w*+c-'', c=getchar(); return q ? -w : w;
 }

 void init()
 {
     cin>>n>>m;
     up=(llg)floor(log2((long double)n));
     for (llg i=;i<=up;i++) for (llg j=;j<=n;j++) for (llg k=;k<=n;k++) dp[i][j][k]=-*inf;
     for (llg s=;s<=up;s++)  for (llg i=;i<=n;i++) dp[s][i][i]=;
     for (llg i=;i<=m;i++)
     {
         llg u=getint(),v=getint();
         dp[][u][v]=getint(),dp[][v][u]=getint();
     }
 }

 void make_dp()
 {
     for (llg s=;s<=up;s++)
         for (llg k=;k<=n;k++)
             for (llg i=;i<=n;i++)
             {
                  if(dp[s - ][i][k]==-inf) continue;
                 for (llg j=;j<=n;j++)
                     dp[s][i][j]=max(dp[s][i][j],dp[s-][i][k]+dp[s-][k][j]);
             }
 }

 bool check(llg x)
 {
     for (llg i=;i<=up;i++) for (llg j=;j<=n;j++) for (llg k=;k<=n;k++) a[i][j][k]=-*inf;
     for (llg i=;i<=n;i++) a[][i][i]=;
     llg p=;
     for (llg s=;s<=up;s++)
     {
         if (((x>>s)&)==) continue;
         p++;
         for (llg i=;i<=n;i++) a[p][i][i]=;
         for (llg k=;k<=n;k++)
             for (llg i=;i<=n;i++)
                 for (llg j=;j<=n;j++)
                     a[p][i][j]=max(a[p][i][j],a[p-][i][k]+dp[s][k][j]);
     }
     for (llg i=;i<=n;i++) if (a[p][i][i]>) return ;
     return ;
 }

 int main()
 {
     yyj("a");
     init();
     make_dp();
     llg l=,r=n+,ans=;
     while (l<=r)
     {
         llg mid=(l+r)>>;
         if (check(mid)) r=mid-,ans=mid;else l=mid+;
     }
     if (l>=n+) ans=;
     cout<<ans;
     return ;
 }