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：一道挺典型的拓扑排序，先把所有入度为0的点入列，进入循环，每次弹出一个节点，找出与这个节点相连的节点，入度减1，更新最小花费时间，再把所有入度为0的节点入列，进入下一次循环。加了个cnt判断有没有环。

#include <iostream>
#include <algorithm>
#include <queue> 
#define MAXN 105
#define INF 1e7
using namespace std;
int Graph[MAXN][MAXN];//矩阵Graph[i][j]代表i->j的cost
void TopSort(int  N) {
	int cnt = 0;
	//计数器判断有无环路
	int* Earliest = new int[N];			//节点的时间花费
	int* Indegree = new int[N];			//入度
	for (int i = 0; i < N; i++) {
		Indegree[i] = 0;				
		Earliest[i] = 0;				
		for (int j = 0; j < N; j++) {	//初始化 入度
			if (Graph[j][i] != INF) {
				Indegree[i]++;
			}
		}
	}
	queue<int> Q;
	for (int i = 0; i < N; i++) {	//起始判断入度为0的节点，放入队列（其实这里用堆栈也可以
		if (Indegree[i] == 0) {
			Q.push(i);
			Indegree[i] = -1;		//标记节点i避免重复入队
			Earliest[i] = 0;		
		}
	}
	while (!Q.empty()) {
		int temp = Q.front();
		Q.pop();
		cnt++;						//处理了一个节点cnt++
		int longest = 0;
		for (int i = 0; i < N; i++) {		//遍历所有节点
			if (Graph[temp][i] != INF) {	//如果有边<temp,i>
				Indegree[i]--;				//i的的入度减1
				Earliest[i] = max(Earliest[i], Earliest[temp] + Graph[temp][i]);//节点i的最长花费进行更新
			}
		}
		for (int i = 0; i < N; i++) {//再次循环判断有没有入度为0的节点，如果有就放入队列
			if (Indegree[i] == 0) {
				Q.push(i);
				Indegree[i] = -1;	 //标记节点避免重复入队
			}
		}
	}
	sort(Earliest, Earliest + N);	
	//对所有节点的时长花费进行从小到大的排序，也处理了有多个终点的情况
	if (cnt == N) {					//如果处理了N个节点
		cout << Earliest[N - 1];	//输出最长的时间花费
	}
	else {
		cout << "Impossible";		//否则代表图中有环
	}
}
int main() {
	int N, M, v1, v2, cost;//节点个数N	边数M	
	cin >> N >> M;
	for (int i = 0; i < N; i++) {		//初始化整张图INF
		for (int j = 0; j < N; j++) {
			Graph[i][j] = INF;
		}
	}
	for (int i = 0; i < M; i++) {		//输入<v1,v2>的cost
		cin >> v1 >> v2 >> cost;
		Graph[v1][v2] = cost;
	}
	TopSort(N);							
}