POJ

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 profitV_i of all cities he may reach (a negative V_i 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 cityi, V_i (0 ≤ |V_i| ≤ 20000)
The next m lines each contain two integers x, y indicating that there is a road leads from cityx 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

题意：一个人去找工作遇到了一道面试题。面试官要求给出一些城市和城市之间的道路，每到达一个城市。可能会赚一些钱，可是也可能会有损失。

终于面试者的所得会决定他能否得到这份工作。显而易见，越多越好。

思路：由于是有向无环图（DAG）并且事实上求的是从一个0入度到0出度的路径，所以我们能够用topsort来处理，再加上简单的DP 即可了

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;
const int maxn = 100005;
const int inf = 0x3f3f3f3f;

struct Node {
	int v, next;
}node[maxn*20];
int n, m, cnt;
int profit[maxn];
int ind[maxn], out[maxn], dp[maxn], adj[maxn];

void topsort() {
	queue<int> q;
	for (int i = 1; i <= n; i++) 
		if (ind[i] == 0) {
			q.push(i);
			dp[i] = profit[i];
		}
	while (!q.empty()) {
		int cur = q.front();
		q.pop();
		for (int i = adj[cur]; i != -1; i = node[i].next) {
			int v = node[i].v;
			if (dp[v] < dp[cur]+profit[v])
				dp[v] = dp[cur]+profit[v];
			if (--ind[v] == 0)
				q.push(v);
		}
	}
}

int main() {
	while (scanf("%d%d", &n, &m) != EOF) {
		cnt = 0;
		memset(adj, -1, sizeof(adj));
		memset(ind, 0, sizeof(ind));
		memset(out, 0, sizeof(out));
		for (int i = 1; i <= n; i++) {
			dp[i] = -inf;
			scanf("%d", &profit[i]);
		}
		for (int i = 0; i < m; i++) {
			int a, b;
			scanf("%d%d", &a, &b);
			out[a]++;
			ind[b]++;
			node[cnt].v = b;
			node[cnt].next = adj[a];
			adj[a] = cnt++;
		}
		topsort();
		int ans = -inf;
		for (int i = 1; i <= n; i++) 
			if (out[i] == 0 && dp[i] > ans)
				ans = dp[i];
		printf("%d
", ans);
	}
	return 0;
}