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06-4. How Long Does It Take (25)
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 4Sample Output 1:
18Sample Input 2:
4 5 0 1 1 0 2 2 2 1 3 1 3 4 3 2 5Sample 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 elseputs("Impossible");return 0;
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