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[21][maxn][maxn],dp[21][maxn][maxn],up;

inline int getint()
{
    int w=0,q=0; char c=getchar();
    while((c<'0' || c>'9') && c!='-') c=getchar(); if(c=='-') q=1,c=getchar(); 
    while (c>='0' && c<='9') w=w*10+c-'0', c=getchar(); return q ? -w : w;
}

void init()
{
    cin>>n>>m;
    up=(llg)floor(log2((long double)n));
    for (llg i=0;i<=up;i++) for (llg j=1;j<=n;j++) for (llg k=1;k<=n;k++) dp[i][j][k]=-1*inf;
    for (llg s=0;s<=up;s++)  for (llg i=1;i<=n;i++) dp[s][i][i]=0;
    for (llg i=1;i<=m;i++)
    {
        llg u=getint(),v=getint();
        dp[0][u][v]=getint(),dp[0][v][u]=getint();
    }
}

void make_dp()
{
    for (llg s=1;s<=up;s++)
        for (llg k=1;k<=n;k++)
            for (llg i=1;i<=n;i++)
            {
                 if(dp[s - 1][i][k]==-inf) continue;
                for (llg j=1;j<=n;j++) 
                    dp[s][i][j]=max(dp[s][i][j],dp[s-1][i][k]+dp[s-1][k][j]);
            }
}

bool check(llg x)
{
    for (llg i=0;i<=up;i++) for (llg j=1;j<=n;j++) for (llg k=1;k<=n;k++) a[i][j][k]=-1*inf;
    for (llg i=1;i<=n;i++) a[0][i][i]=0;
    llg p=0;
    for (llg s=0;s<=up;s++)
    {
        if (((x>>s)&1)==0) continue;
        p++;
        for (llg i=1;i<=n;i++) a[p][i][i]=0;
        for (llg k=1;k<=n;k++)
            for (llg i=1;i<=n;i++)
                for (llg j=1;j<=n;j++)
                    a[p][i][j]=max(a[p][i][j],a[p-1][i][k]+dp[s][k][j]);
    }
    for (llg i=1;i<=n;i++) if (a[p][i][i]>0) return 1;
    return 0;
}

int main()
{
    yyj("a");
    init();
    make_dp();
    llg l=2,r=n+1,ans=0;
    while (l<=r)
    {
        llg mid=(l+r)>>1;
        if (check(mid)) r=mid-1,ans=mid;else l=mid+1;
    }
    if (l>=n+1) ans=0;
    cout<<ans;
    return 0;
}