mooc -- 06-4 How Long Does It Take

题目：

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

这道题就是一道拓扑排序的题求最短完成时间

注意：

输出最大值，可以用标准库的函数：max_element，返回值是地址，所以需要*把内容取出来。如数组a
，最大值为：*max_element(earlist, earlist+n)。当然不是求数组的最大值，而是求vector的最大值，只需要传相应的迭代器就可以。

代码中都有说明，直接上：

#include<iostream>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;

struct node       //定义结构结点
{
int S;
int E;
int L;
node(int a,int b,int c):S(a),E(b),L(c) {}
} ;

//cmp函数（用于sort排序函数的子函数）
bool cmp(const node &a, const node &b)
{
return a.E < b.E;
}

int main()
{
//最早完成时间、入度数组的动态申请及初始化
int N,M;
cin>>N>>M;
int *earlist = new int
;
int *Indegree = new int
;

for(int i=0;i<N;i++)
{
earlist[i] = 0;
Indegree[i] = 0;
}

vector<node> vec;        //定义一个容器存放入度为0 的结点
//s，e输入并存储
for(int i=0;i<M;i++)
{
int s,e,l;
cin>>s>>e>>l;
Indegree[e] ++ ;
vec.push_back(node(s,e,l));
}
//sort函数用于排序，把终点相同的活动集中在一起
sort(vec.begin(), vec.end(), cmp);

queue<int> q;

for(int v=0;v<N;v++)
{
if(Indegree[v] == 0)
q.push(v);
}
//拓扑排序
int cnt = 0;

while( q.size() != 0)
{
int v = q.front();
q.pop();
cnt++;
for ( int i=0; i<M; i++ )
{
if ( vec[i].S == v)
{
int w = vec[i].E;
earlist[w] = max(earlist[w], earlist[v] + vec[i].L);
if ( --Indegree[w] == 0)
q.push(w);
}
}

}
//判断是否存在回路
if(cnt != N)
cout<<"Impossible"<<endl;
else
cout<<*max_element(earlist, earlist+N)<<endl;
return 0;
}