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 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

#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;
        }
    }
}

标签：25,cur,map,int,inDegree,08,Long,++,maxn
来源： https://www.cnblogs.com/wanghao-boke/p/11827748.html