这道题的大意就是，给出一个有向图，每个点有一个点权，点权可能是正也可能为负，一个人从某点出发，沿着一些路，访问结点，或者仅仅是路过这个结点，而不去访问，最后求他能访问到的最大的点权和。
我们注意到，他对某个结点是可以选择访问或者不访问的，那么只用访问那些点权为正数的点了。
首先，求强连通分量，缩点，然后新点的点权就是原强连通分量中，所有正点权之和。
之后就要进行DP求解，我在本题中使用了Kosaraju算法，其好处就是，最后缩点后，恰好形成拓扑序列，而DP就要从拓扑序靠后的点往前进行DP，这样才能从叶子结点往根部进行DP，从而得到正确答案
/*
ID: sdj22251
PROG: subset
LANG: C++
*/
#include <iostream>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cmath>
#include <ctime>
#define LOCA
#define MAXN 5005
#define INF 100000000
#define eps 1e-7
using namespace std;
struct Edge
{int v, next;
}edge[10 * MAXN], revedge[10 * MAXN], newedge[MAXN];
int head[MAXN], revhead[MAXN], e, visited[MAXN], newhead[MAXN];
int order[MAXN], cnt, id[MAXN], val[MAXN], newval[MAXN];
int uu[5 * MAXN], vv[5 * MAXN], out[MAXN], dp[MAXN];
int n, m;
int ans;
void init()
{e = 0;memset(head, -1, sizeof(head));memset(revhead, -1, sizeof(revhead));memset(newhead, -1, sizeof(newhead));memset(out, 0 , sizeof(out));memset(newval, 0, sizeof(newval));memset(dp, 0, sizeof(dp));
}
void insert(const int &x, const int &y)
{edge[e].v = y;edge[e].next = head[x];head[x] = e;revedge[e].v = x;revedge[e].next = revhead[y];revhead[y] = e;e++;
}
void newinsert(const int &x, const int &y)
{newedge[e].v = y;newedge[e].next = newhead[x];newhead[x] = e++;
}
void readdata()
{for(int i = 1; i <= n; i++)scanf("%d", &val[i]);for(int i = 0; i < m; i++){scanf("%d%d", &uu[i], &vv[i]);uu[i]++;vv[i]++;insert(uu[i], vv[i]);}
}
void dfs(int u)
{visited[u] = 1;for(int i = head[u]; i != -1; i = edge[i].next){int v = edge[i].v;if(!visited[v])dfs(v);}order[cnt++] = u;
}
void dfs_rev(int u)
{visited[u] = 1;id[u] = cnt;if(val[u] > 0) newval[cnt] += val[u];for(int i = revhead[u]; i != -1; i = revedge[i].next){int v = revedge[i].v;if(!visited[v])dfs_rev(v);}
}
void Kosaraju()
{init();readdata();memset(visited, 0, sizeof(visited));cnt = 0;for(int i = 1; i <= n; i++){if(!visited[i])dfs(i);}memset(visited, 0, sizeof(visited));cnt = 0;for(int i = n - 1; i >= 0; i--){if(!visited[order[i]]){cnt++;dfs_rev(order[i]);}}e = 0;for(int i = 0; i < m; i++){int u = id[uu[i]];int v = id[vv[i]];if(u != v) newinsert(u, v);}ans = 0;for(int u = cnt; u >= 1; u--){int tmp = 0;dp[u] = newval[u];for(int i = newhead[u]; i != -1; i = newedge[i].next){int v = newedge[i].v;tmp = max(tmp, dp[v]);}dp[u] += tmp;ans = max(ans, dp[u]);}
}
int main()
{while(scanf("%d%d", &n, &m) != EOF){Kosaraju();printf("%d\n", ans);}return 0;
}