可耻的苟到数据后搞了一波题

首先标解绝对不是倍增floyd

这是n^3logn的算法

二分+DFS_SPFA是nlogn的算法

但是在学习嘛

所谓标解：

#include<bits/stdc++.h>
using namespace std;
const int N=310;
int n,m;
struct Matrix{int a[N][N];friend Matrix operator * (Matrix A,Matrix B){Matrix C;memset(C.a,0x3f,sizeof(C.a));for(int k=1;k<=n;k++){for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){C.a[i][j]=min(C.a[i][j],A.a[i][k]+B.a[k][j]);}}}return C;}
}a[N],x;
int main(){
//  cout<<"218"<<'\n';scanf("%d%d",&n,&m);
//  cout<<"8791897"<<'\n';memset(x.a,0x3f,sizeof(x.a));memset(a[0].a,0x3f,sizeof(a[0].a));for(int i=1;i<=m;i++){int u,v,w;scanf("%d%d%d",&u,&v,&w);a[0].a[u][v]=w;    }for(int i=1;i<=n;i++){a[0].a[i][i]=x.a[i][i]=0;}int len=0;while((1<<len)<=n)len++;len--;for(int i=1;i<=len;i++){a[i]=a[i-1]*a[i-1];}Matrix tmp;int ans=0;for(int j=len;j>=0;j--){tmp=x*a[j];int flag=1;for(int i=1;i<=n;i++){if(tmp.a[i][i]<0){flag=0;break;  }       }if(flag==1)x=tmp,ans+=(1<<j);}if(ans>n)cout<<0;cout<<ans+1;
} 

比标解快几倍的解法：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
using namespace std;const int ONE = 305;
const int EDG = ONE*ONE; int n,m;
int x,y,z;
int nxt[EDG],first[EDG],go[EDG],w[EDG],tot;
int vis[ONE],dist[ONE];
int PD;int get()
{int res=1,Q=1;    char c;while( (c=getchar())<48 || c>57)if(c=='-')Q=-1;if(Q) res=c-48; while((c=getchar())>=48 && c<=57) res=res*10+c-48; return res*Q;
}void Add(int u,int v,int z)
{nxt[++tot]=first[u];    first[u]=tot;    go[tot]=v;    w[tot]=z;
}void Spfa(int u,int T,int Limit)
{if(PD) return;for(int e=first[u];e;e=nxt[e]){int v = go[e];if(dist[u]+w[e] <= dist[v]){if(vis[v]) {PD = 1; return;}if(T==Limit) return;dist[v] = dist[u] + w[e];vis[v] = 1;Spfa(v,T+1,Limit);vis[v] = 0;}}
}int Check(int Limit)
{PD = 0;for(int i=1;i<=n;i++){memset(vis,0,sizeof(vis));  vis[i] = 1;memset(dist,0,sizeof(dist));Spfa(i,1,Limit);if(PD) return 1;}return 0;
}int main()
{n=get();    m=get();for(int i=1;i<=m;i++){x=get();    y=get();    z=get();Add(x,y,z);}if(!Check(n)) {printf("0"); exit(0);}int l=1, r=n;while(l < r-1){int mid = l+r>>1;if(Check(mid)) r = mid;else l = mid;}if(Check(l)) printf("%d",l);else printf("%d",r);
}