08-图8 How Long Does It Take（25 分）

Given the relations of all the activities of a project, you are supposed to find the earliest completion time of the project.

Input Specification:
Each input file contains one test case. Each case starts with a line containing two positive integers N (≤100), the number of activity check points (hence it is assumed that the check points are numbered from 0 to N−1), and M, the number of activities. Then M lines follow, each gives the description of an activity. For the i-th activity, three non-negative numbers are given: S[i], E[i], and L[i], where S[i] is the index of the starting check point, E[i] of the ending check point, and L[i] the lasting time of the activity. The numbers in a line are separated by a space.

Output Specification:
For each test case, if the scheduling is possible, print in a line its earliest completion time; or simply output “Impossible”.

Sample Input 1:
9 12
0 1 6
0 2 4
0 3 5
1 4 1
2 4 1
3 5 2
5 4 0
4 6 9
4 7 7
5 7 4
6 8 2
7 8 4
Sample Output 1:
18
Sample Input 2:
4 5
0 1 1
0 2 2
2 1 3
1 3 4
3 2 5
Sample Output 2:
Impossible

###思路：
关键路径问题，以 dist[i] 数组记录到 i 节点的最短时间。以 inDegree[i] 数组记录 i 节点的入度。
每次把入度为零的节点入队，入队之后将该节点指向的节点入度减一。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<queue>
using namespace std;
const int N = 102;
queue<int> q;
int graph[N][N] = { 0 };
int inDegree[N] = { 0 };
int dist[N];
bool topsort(int n) {
	//初始化dist 
	for (int i = 0; i < n; i++) {
		dist[i] = -1;
	}
	int cur = 0;
	for (int i = 0; i < n; i++) {
		if (inDegree[i] == 0) {
			q.push(i);
			dist[i] = 0;
			for (int j = 0; j < n; j++) {
				if (graph[i][j] > 0) {
					dist[j] = graph[i][j];
				}
			}	
		}		
	}

	int cnt = 0;
	while (!q.empty()) {
		cur = q.front();
		q.pop();
		cnt++;
		for (int i = 0; i < n; i++) {
			if (graph[cur][i] >= 0) {
				inDegree[i]--;
				if (dist[cur] + graph[cur][i] > dist[i]) {
					dist[i] = graph[cur][i] + dist[cur];
				}
				if (inDegree[i] == 0) {
					q.push(i);
				}
			}
		}
	}
	if (cnt != n) {
		return false;
	}
	return true;
}
int main() {
	int n, m;
	cin >> n >> m;
	for (int i = 0;  i < n; i++) {
		for (int j = 0; j < n; j++) {
			graph[i][j] = -1;
		}
	}
	while (m--) {
		int s, e, l;
		cin >> s >> e >> l;
		graph[s][e] = l;
		inDegree[e]++;
	}
	
	if (!topsort(n))
		cout << "Impossible" << endl;
	else {
		int max = -1;
		for (int i = 0; i < n; i++) {
			max = max > dist[i] ? max : dist[i];
		}
		cout << max << endl;
	}
	return 0;
}