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[105][105],dist[105],indegree[105];
int main(){
    //freopen("D:\INPUT.txt","r",stdin);
    int n,m;
    int i,j,a,b,cost;
    scanf("%d %d",&n,&m);
    for(i=0;i<n;i++){
        dist[i]=-1;
        indegree[i]=0;
        for(j=0;j<n;j++){
            Map[i][j]=Map[j][i]=-1;
        }
    }
    for(i=0;i<m;i++){
        scanf("%d %d %d",&a,&b,&cost);
        Map[a][b]=cost;
        indegree[b]++;
    }
    queue<int> q;
    for(i=0;i<n;i++){
        if(!indegree[i]){
            q.push(i);
            dist[i]=0;//这个初始化很重要
        }
    }
    int cur;
    while(!q.empty()){
        cur=q.front();
        q.pop();
        for(i=0;i<n;i++){
            if(Map[cur][i]!=-1){
               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=-1;
    for(i=0;i<n;i++){
        if(indegree[i]){
            break;
        }
        if(dist[i]>maxcost){
            maxcost=dist[i];
        }
    }
    if(i==n){
        printf("%d
",maxcost);
    }
    else{
        printf("Impossible
");
    }
    return 0;
}