zoukankan      html  css  js  c++  java
  • HDU 2647 Reward(toposort)

    Reward




    Problem Description
    Dandelion's uncle is a boss of a factory. As the spring festival is coming , he wants to distribute rewards to his workers. Now he has a trouble about how to distribute the rewards.
    The workers will compare their rewards ,and some one may have demands of the distributing of rewards ,just like a's reward should more than b's.Dandelion's unclue wants to fulfill all the demands, of course ,he wants to use the least money.Every work's reward will be at least 888 , because it's a lucky number.
     
    Input
    One line with two integers n and m ,stands for the number of works and the number of demands .(n<=10000,m<=20000)
    then m lines ,each line contains two integers a and b ,stands for a's reward should be more than b's.
     
    Output
    For every case ,print the least money dandelion 's uncle needs to distribute .If it's impossible to fulfill all the works' demands ,print -1.
     
    Sample Input
    2 1
    1 2
    2 2
    1 2
    2 1
     
    Sample Output
    1777
    -1
     
    建反向图编号
     
     1 #include<cstdio>
     2 #include<cstring>
     3 #include<vector>
     4 using namespace std;
     5 
     6 const int maxn=10005;
     7 
     8 struct Node
     9 {
    10     int c;
    11     int id;
    12 }node[maxn];
    13 
    14 vector<int>G[maxn];
    15 bool indegree[maxn];
    16 int n,m;
    17 
    18 void init()
    19 {
    20     for(int i=1;i<=n;i++)
    21         G[i].clear();
    22     memset(node,0,sizeof(node));
    23     memset(indegree,0,sizeof(indegree));
    24 }
    25 
    26 bool dfs(int u,int id)
    27 {
    28     node[u].c=-1;
    29     for(int i=0;i<G[u].size();i++)
    30         if(node[G[u][i]].c<0)return false;
    31     else if(!dfs(G[u][i],id+1))return false;
    32         node[u].c=1;
    33         node[u].id=max(node[u].id,id);
    34     return true;
    35 }
    36 
    37 bool toposort()
    38 {
    39     for(int i=1;i<=n;i++)
    40         if(!node[i].c&&!indegree[i])
    41         {
    42             if(!dfs(i,0))
    43             return false;
    44         }
    45     for(int i=1;i<=n;i++)
    46         if(!node[i].c)
    47         return false;
    48     return true;
    49 }
    50 
    51 int main()
    52 {
    53     int a,b;
    54     while(scanf("%d%d",&n,&m)!=EOF)
    55     {
    56         init();
    57         for(int i=0;i<m;i++)
    58         {
    59              scanf("%d%d",&a,&b);
    60              G[b].push_back(a);
    61              indegree[a]=true;
    62         }
    63         bool ans=toposort();
    64         if(ans)
    65         {
    66             int temp=0;
    67             for(int i=1;i<=n;i++)
    68                 temp+=node[i].id;
    69             printf("%d
    ",888*n+temp);
    70         }
    71         else
    72             printf("-1
    ");
    73     }
    74     return 0;
    75 }
  • 相关阅读:
    2018年春季个人阅读计划
    软件需求与分析读后感
    假期读后感3
    假期读后感2
    假期读后感1
    四则运算2
    软件工程概论第一次作业
    《大道至简》读后感
    HMX-Server C++ 分步式服务器大版本更新了(有源码)
    HMX-Server-分步式服务器框架(开源+源码)
  • 原文地址:https://www.cnblogs.com/homura/p/4729647.html
Copyright © 2011-2022 走看看