zoukankan      html  css  js  c++  java
  • 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 N1), 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;
            }
        }
    }
     
  • 相关阅读:
    opengl中对glOrtho()函数的理解
    cocos2D-x demo 的源码分析 #define ..##.. 的妙用.
    js练习图片轮播
    js 表单操作form
    JS DOM
    java Map
    html--form表单
    java反射应用
    JDBC的使用-----Statement
    sql 查询语句的练习2
  • 原文地址:https://www.cnblogs.com/wanghao-boke/p/11827748.html
Copyright © 2011-2022 走看看