pat06-图7. How Long Does It Take (25)

06-图7. 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

---

提交代码

AOV图。拓扑排序。

 #include<cstdio>
 #include<algorithm>
 #include<iostream>
 #include<cstring>
 #include<queue>
 #include<vector>
 #include<cmath>
 #include<string>
 using namespace std;
 int Map[][],dist[],indegree[];
 int main(){
     //freopen("D:\\INPUT.txt","r",stdin);
     int n,m;
     int i,j,a,b,cost;
     scanf("%d %d",&n,&m);
     for(i=;i<n;i++){
         dist[i]=-;
         indegree[i]=;
         for(j=;j<n;j++){
             Map[i][j]=Map[j][i]=-;
         }
     }
     for(i=;i<m;i++){
         scanf("%d %d %d",&a,&b,&cost);
         Map[a][b]=cost;
         indegree[b]++;
     }
     queue<int> q;
     for(i=;i<n;i++){
         if(!indegree[i]){
             q.push(i);
             dist[i]=;//这个初始化很重要
         }
     }
     int cur;
     while(!q.empty()){
         cur=q.front();
         q.pop();
         for(i=;i<n;i++){
             if(Map[cur][i]!=-){
                indegree[i]--;
                if(dist[cur]+Map[cur][i]>dist[i]){
                     dist[i]=dist[cur]+Map[cur][i];
                 }
                if(!indegree[i]){
                     q.push(i);
                }
             }
         }
     }
     int maxcost=-;
     for(i=;i<n;i++){
         if(indegree[i]){
             break;
         }
         if(dist[i]>maxcost){
             maxcost=dist[i];
         }
     }
     if(i==n){
         printf("%d\n",maxcost);
     }
     else{
         printf("Impossible\n");
     }
     return ;
 }