06-4. How Long Does It Take (25)

时间限制 200 ms

内存限制 65536 kB

代码长度限制 8000 B

判题程序
Standard 作者 CHEN, Yue

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

#include <iostream>
#include <algorithm>
#include <string>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include<queue>
#include<stack>
#include<map>
#include<set>
using namespace std;
#define lson rt<<1,l,MID
#define rson rt<<1|1,MID+1,r
//#define lson root<<1
//#define rson root<<1|1
#define MID ((l+r)>>1)
typedef long long ll;
typedef pair<int,int> P;
const int maxn=3005;
const int base=1000;
const int inf=9999999999;
const double eps=1e-5;
int n,m;
struct node
{
    int to,cost;
};
vector<node> G[maxn];//邻接表
int d[maxn];
int earlist[maxn];//活动最早结束的数组
bool tupo_sort()//拓扑排序 求最长的路径
{
    stack<int> s;
    for(int i=0;i<n;i++)
    {
        if(d[i]==0)s.push(i);
    }
    int cnt=0;//用于判断是否有环
    while(!s.empty())
    {
        int m=s.top();
        cnt++;s.pop();
        for(int i=0;i<G[m].size();i++)
        {
            node tmp=G[m][i];
            if(--d[tmp.to]==0)s.push(tmp.to);
            if(earlist[m]+tmp.cost>earlist[tmp.to])//计算最长的路径
            {
                earlist[tmp.to]=earlist[m]+tmp.cost;
            }
        }
    }
    if(cnt<n)return false;//判断是否有环
    return true;
}
int main()
{
    int i,j,k,t;
    cin>>n>>m;
    for(i=0;i<m;i++)
    {
        node e;
        int s;
        cin>>s>>e.to>>e.cost;
        G[s].push_back(e);
        d[e.to]++;
    }
    if(tupo_sort())
        printf("%d\n",*max_element(earlist,earlist+n));//输出earlist数组的最大元素 
    else
         puts("Impossible");

    return 0;
}