[题解] poj 3249 Test for Job (拓扑排序 + DP)

- 传送门 -

　http://poj.org/problem?id=3249

#Test for Job

| Time Limit: 5000MS |   | Memory Limit: 65536K |
| Total Submissions: 11340 |   | Accepted: 2680 |

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 V_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 _i_th line describes the net profit of the city i, V_i (0 ≤ |V_i| ≤ 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

Source

POJ Monthly--2007.07.08, 落叶飞雪

- 题意 -

　给 n 个有权值的点, 一些有向边, 求从一个入度为 0 的点走到一个出度为 0 的点的最大权值合.
　

- 思路 -

　dp + (被我写得像暴力的)拓扑排序.
　入度为 0 的点赋初值, 出度为 0 的点求答案.
　
　细节见代码.
　

- 代码 -

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;

const int N = 1e5 + 5;
const int M = 1e6 + 5;
const int inf = 0x3f3f3f3f;

int TO[M], NXT[M], HD[N];
int IN[N], OUT[N], DP[N];
int COST[N];
int n, m, x, a, b, sz;
queue<int>q;

void topo() {
	while (!q.empty())
		q.pop();
	for (int i = 1; i <= n; ++i) {
		if (IN[i] == 0) {
			q.push(i);
			DP[i] = COST[i];
			IN[i] = -1;
		}
	}
	while (!q.empty()) {
		int u = q.front();
		q.pop();
		for (int i = HD[u]; i; i = NXT[i]) {
			int v = TO[i];
			if (IN[v] > 0) {
				IN[v]--;
				if (DP[v] < DP[u] + COST[v])
					DP[v] = DP[u] + COST[v];
				if (IN[v] == 0) {
					q.push(v);
					IN[v] --;
				}
			}
		}
	}
}

int main() {
	while (~scanf("%d%d", &n, &m)) {
		sz = 0;
		for (int i = 1; i <= n; ++i) {
			scanf("%d", &COST[i]);
			HD[i] = IN[i] = OUT[i] = 0;
			DP[i] = -inf;
		}
		for (int i = 1; i <= m; ++i) {
			scanf("%d%d", &a, &b);
			OUT[a] ++;
			IN[b] ++;
			NXT[++sz] = HD[a];
			HD[a] = sz;
			TO[sz] = b;
		}
		topo();
		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;
}