我的思路是先tarjan判环，然后缩点。最后以所有入度为0的点为起点，spfa跑最长路，最后不断对ans取max

#include<bits/stdc++.h>
#define ll long long
#define lowbit(x) x&(-x)
using namespace std;
const int N=1e4+10;
inline int read()
{
	ll x=0,f=1;char ch=getchar();
	while (!isdigit(ch)){if (ch=='-') f=-1;ch=getchar();}
	while (isdigit(ch)){x=x*10+ch-48;ch=getchar();}
	return x*f;
}
int n,m,s;
vector<int>g[N],g2[N];
int dfn[N],low[N],id;
bool vis[N];
int stk[N],top;
int co[N],tot;
int a[N];
int sum[N];
void tarjan(int u){
	dfn[u]=low[u]=++id;
	vis[u]=1;
	stk[++top]=u;
	for(int i=0;i<g[u].size();++i){
		int v=g[u][i];
		if(!dfn[v]){
			tarjan(v);
			low[u]=min(low[u],low[v]);
		}
		else if(vis[v])low[u]=min(low[u],dfn[v]);
	}
	if(dfn[u]==low[u]){
		tot++;
		while(stk[top]!=u){
			co[stk[top]]=tot;
			vis[stk[top]]=0;
			top--;
		}
		co[u]=tot;
		vis[u]=0;
		top--;
	}
}
bool inq[N];
int dis[N];
int ans=INT_MIN;
void spfa(int now){
	queue<int>q;
	memset(dis,-0x3f,sizeof(dis));
	memset(inq,0,sizeof inq);
	dis[now]=sum[now];
	q.push(now);
	inq[now]=1;
	while(!q.empty()){
		int u=q.front();
		q.pop();
		inq[now]=0;
		for(int i=0;i<g2[u].size();++i){
			int v=g2[u][i];
			if(inq[v])continue;
			if(dis[v]<dis[u]+sum[v]){
				dis[v]=dis[u]+sum[v];
				ans=max(ans,dis[v]);
				q.push(v);
				inq[v]=1;
			}
			
		}
	}
}
int in[N];
int main()
{
	n=read(),m=read();
	for(int i=1;i<=n;++i)a[i]=read();
	for(int i=1;i<=m;++i){
		int a=read(),b=read();
		g[a].push_back(b);
	}
	for(int i=1;i<=n;++i)if(!dfn[i])tarjan(i);
	for(int i=1;i<=n;++i){
		for(int j=0;j<g[i].size();++j){
			int v=g[i][j];
			if(co[v]==co[i]){//i,v在一个环内；
				sum[co[i]]+=a[v];
			}
			else g2[co[i]].push_back(co[v]),in[co[v]]++;
		}
	}
	for(int i=1;i<=tot;++i){
		if(!in[i])spfa(i);
	}
//	cout<<endl;
	cout<<ans;
}