7-11 How Long Does It Take（25 分）（toposort）

7-11 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 $N$ ($\le 100$), 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

ac:代码

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
const int maxn = 505;
int map[maxn][maxn],dis[maxn];//map有向图dis距离 
int in[maxn];//记录入度 
int n,m;
int toposort()
{
	queue<int>que;
	memset(dis,0,sizeof(dis));
	int cnt = 0;
	for(int i = 0;i<n;i++){//选取入度为零的点，入队列 
		if(in[i]==0){
			que.push(i);
		}
	}
	while(!que.empty()){
		int v = que.front();//取对顶 
		cnt++;
		que.pop();
		for(int i = 0;i<n;i++){
			if(map[v][i]!=-1){//如果这条边存在，则更新长度 
				dis[i] = max(dis[i],map[v][i] + dis[v]);
				if(--in[i]==0) que.push(i);
			}
		}
	}
	if(cnt==n) return 0;
	else return 1;
}
int main()
{
	memset(dis,0,sizeof(dis));
	memset(in,0,sizeof(in));
	memset(map,-1,sizeof(map));
	cin>>n>>m;
	int from,to,cost;
	for(int i = 0;i<m;i++)
	{
		scanf("%d %d %d",&from,&to,&cost);
		map[from][to] = cost;//有向图的赋值 
		in[to]++;
	}
	int ans = toposort();
	if(ans) printf("Impossible
");
	else{
		int max1 = -1;
		for(int i = 0;i<n;i++){
			if(max1<dis[i]){
				max1 = dis[i];
			}
			
		}
		printf("%d",max1);
	}
	
	return 0;
}

现在我来解释一下入度，如上图（手绘图），没有箭头指向1因此1的入度为零，同理3的入度也为零，2的入度为1,4的入度为2.