7-11 How Long Does It Take（25 分）

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7-11 How Long Does It Take（25 分）

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 $N$ ($\le 100$), 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

拓扑排序

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
int n,m;
int in[110],map[110][110],cost[110];
int main()
{
int i,j,a,b,u,max;
scanf("%d %d",&n,&m);
for(i=0;i<n;i++)
{
in[i]=0;
cost[i]=-1;
for(j=i;j<n;j++)
map[i][j]=map[j][i]=-1;
}
queue<int>q;
while(m--)
{
scanf("%d%d%d",&a,&b,&u);
in[b]++;
map[a][b]=u;
}
for(i=0;i<n;i++)
if(in[i]==0)
{
q.push(i);
cost[i]=0;
}
while(!q.empty())
{
u=q.front();
q.pop();
for(i=0;i<n;i++)
{
if(map[u][i]!=-1)
{
in[i]--;
if(cost[i]<cost[u]+map[u][i])
cost[i]=cost[u]+map[u][i];
if(in[i]==0)
q.push(i);
}
}
}
max=-1;
for(i=0;i<n;i++)
{
if(in[i])//成环 
break;
if(cost[i]>max)
max=cost[i];
}
if(i==n)
printf("%d\n",max);
else
printf("Impossible\n");
return 0;
}