08-圖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

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：一道挺典型的拓撲排序，先把所有入度為0的點入列，進入迴圈，每次彈出一個節點，找出與這個節點相連的節點，入度減1，更新最小花費時間，再把所有入度為0的節點入列，進入下一次迴圈。加了個cnt判斷有沒有環。

#include <iostream>
#include <algorithm>
#include <queue> 
#define MAXN 105
#define INF 1e7
using namespace std;
int Graph[MAXN][MAXN];//矩陣Graph[i][j]代表i->j的cost
void TopSort(int  N) {
	int cnt = 0;
	//計數器判斷有無環路
	int* Earliest = new int[N];			//節點的時間花費
	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++) {	//起始判斷入度為0的節點，放入佇列（其實這裡用堆疊也可以
		if (Indegree[i] == 0) {
			Q.push(i);
			Indegree[i] = -1;		//標記節點i避免重複入隊
			Earliest[i] = 0;		
		}
	}
	while (!Q.empty()) {
		int temp = Q.front();
		Q.pop();
		cnt++;						//處理了一個節點cnt++
		int longest = 0;
		for (int i = 0; i < N; i++) {		//遍歷所有節點
			if (Graph[temp][i] != INF) {	//如果有邊<temp,i>
				Indegree[i]--;				//i的的入度減1
				Earliest[i] = max(Earliest[i], Earliest[temp] + Graph[temp][i]);//節點i的最長花費進行更新
			}
		}
		for (int i = 0; i < N; i++) {//再次迴圈判斷有沒有入度為0的節點，如果有就放入佇列
			if (Indegree[i] == 0) {
				Q.push(i);
				Indegree[i] = -1;	 //標記節點避免重複入隊
			}
		}
	}
	sort(Earliest, Earliest + N);	
	//對所有節點的時長花費進行從小到大的排序，也處理了有多個終點的情況
	if (cnt == N) {					//如果處理了N個節點
		cout << Earliest[N - 1];	//輸出最長的時間花費
	}
	else {
		cout << "Impossible";		//否則代表圖中有環
	}
}
int main() {
	int N, M, v1, v2, cost;//節點個數N	邊數M	
	cin >> N >> M;
	for (int i = 0; i < N; i++) {		//初始化整張圖INF
		for (int j = 0; j < N; j++) {
			Graph[i][j] = INF;
		}
	}
	for (int i = 0; i < M; i++) {		//輸入<v1,v2>的cost
		cin >> v1 >> v2 >> cost;
		Graph[v1][v2] = cost;
	}
	TopSort(N);							
}