Description

Mr.Dog was fired by his company. In order to support his family, he must find a new job as soon as possible. Nowadays, It’s hard to have a job, since there are swelling numbers of the unemployed. So some companies often use hard tests for their recruitment.

The test is like this: starting from a source-city, you may pass through some directed roads to reach another city. Each time you reach a city, you can earn some profit or pay some fee, Let this process continue until you reach a target-city. The boss will compute the expense you spent for your trip and the profit you have just obtained. Finally, he will decide whether you can be hired.

In order to get the job, Mr.Dog managed to obtain the knowledge of the net profit Vi of all cities he may reach (a negative Vi indicates that money is spent rather than gained) and the connection between cities. A city with no roads leading to it is a source-city and a city with no roads leading to other cities is a target-city. The mission of Mr.Dog is to start from a source-city and choose a route leading to a target-city through which he can get the maximum profit.

Input

The input file includes several test cases. The first line of each test case contains 2 integers n and m(1 ≤ n ≤ 100000, 0 ≤ m ≤ 1000000) indicating the number of cities and roads. The next n lines each contain a single integer. The ith line describes the net profit of the city i, Vi (0 ≤ |Vi| ≤ 20000) The next m lines each contain two integers x, y indicating that there is a road leads from city x to city y. It is guaranteed that each road appears exactly once, and there is no way to return to a previous city.

Output

The output file contains one line for each test cases, in which contains an integer indicating the maximum profit Dog is able to obtain (or the minimum expenditure to spend)

Sample Input

6 5
1
2
2
3
3
4
1 2
1 3
2 4
3 4
5 6

Sample Output

7

Hint

思路

回顾一下拓扑排序~

先说题意，有一个 $n$ 的节点的 $DAG$ (有向无环)图，每一个节点都有一个权值，现在题目让你求出，从一个入度为0 的点到一个出度为 0 的点的最大点权和。

这是一个 dp ，我们定义 dp[i] 代表满足题目前提的条件下，走到这一点的最大值。易得状态转移方程为：

$$dp[v]=max(dp[v],dp[u]+val[i])$$

那么我们只需要在拓扑排序的时候，进行一下状态转移，最后求出最大值即可。

拓扑排序的办法是找一个入度为 0 的点，删除它的所有出边，可以用一个队列实现。

代码

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#include <cstdio>
#include <queue>
#include <vector>
using namespace std;
const int N = 100000 + 10;
const int inf = 0x3f3f3f3f;
int n, m;
int val[N], in[N], out[N], dp[N];
vector<int> e[N];
void topsort()
{
    queue<int> q;
    for (int i = 1; i <= n; i++)
        if (!in[i])
        {
            dp[i] = val[i];
            q.push(i);
        }
        else
            dp[i] = -inf;
    while (!q.empty())
    {
        int u = q.front();
        q.pop();
        for (int i = 0; i < e[u].size(); i++)
        {
            int v = e[u][i];
            dp[v] = max(dp[v], dp[u] + val[v]);
            in[v]--;
            if (!in[v])
                q.push(v);
        }
    }
}
int main()
{
    //freopen("in.txt", "r", stdin);
    int u, v;
    while (~scanf("%d%d", &n, &m))
    {
        for (int i = 1; i <= n; i++)
        {
            scanf("%d", &val[i]);
            e[i].clear();
            in[i] = out[i] = 0;
        }
        for (int i = 1; i <= m; i++)
        {
            scanf("%d%d", &u, &v);
            e[u].push_back(v);
            in[v]++, out[u]++;
        }
        topsort();
        int ans = -inf;
        for (int i = 1; i <= n; i++)
            if (!out[i] && dp[i] > ans)
                ans = dp[i];
        printf("%d\n", ans);
    }
    return 0;
}

最后修改于 2019-03-03