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• # 08-图8 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

### 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<stdio.h>
#include<queue>
using namespace std;
const int maxn = 110;

int map[maxn][maxn],d[maxn];
int inDegree[maxn];

void init(int n);

int main()
{
int n,m;
scanf("%d%d",&n,&m);

init(n);

int u,v,w;
for (int i = 0; i < m; i++)
{
scanf("%d%d%d",&u,&v,&w);
map[u][v] = w;
inDegree[v]++;
}

queue<int> q;

for (int i  = 0; i < n; i++)
{
if (!inDegree[i])
{
q.push(i);
d[i] = 0;
}
}

while (!q.empty())
{
int cur = q.front();
q.pop();

for (int i = 0; i < n; i++)
{
if (map[cur][i] != -1)
{
inDegree[i]--;
if (d[i] < d[cur] + map[cur][i])
{
d[i] = d[cur] + map[cur][i];
}
if (!inDegree[i])
{
q.push(i);
}
}
}
}

int maxCost = -1;
bool flag = true;
for (int i = 0; i < n; i++)
{
if (inDegree[i])
{
flag = false;
break;
}
if (d[i] > maxCost)
{
maxCost = d[i];
}
}

if (flag)
{
printf("%d",maxCost);
}
else
{
printf("Impossible");
}

return 0;
}

void init(int n)
{
for (int i = 0 ; i < n; i++)
{
d[i] = -1;
inDegree[i] = 0;
for (int j = 0; j < n; j++)
{
map[i][j] = map[j][i] = -1;
}
}
}
 
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