zoukankan      html  css  js  c++  java
  • hdu 2647

    Reward

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 4093    Accepted Submission(s): 1240


    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
    反向建拓扑图,然后设置一个深度数组,用来记录是第几层
     1 #include<string>
     2 #include<cstdio>
     3 #include<iostream>
     4 #include<vector>
     5 #include<queue>
     6 #include<stack>
     7 #include<algorithm>
     8 #include<cstring>
     9 #include<stdlib.h>
    10 #include<string>
    11 #include<cmath>
    12 using namespace std;
    13 #define pb push_back
    14 vector<int >p[10010];
    15 int in[10010],n,m,deep[10010],num[10010];
    16 void init(){
    17     memset(in,0,sizeof(in));
    18     memset(deep,0,sizeof(deep));
    19     memset(num,0,sizeof(num));
    20     for(int i=0;i<=n;i++) p[i].clear();
    21 }
    22 void tuopu(){
    23     queue<int >ak_47;
    24     int cnt=0;
    25     for(int i=1;i<=n;i++)
    26     if(in[i]==0)
    27         ak_47.push(i),deep[i]=1,num[1]++;
    28     while(!ak_47.empty()){
    29         int pos=ak_47.front();
    30         cnt++;
    31         for(int i=0;i<p[pos].size();i++){
    32             int to=p[pos][i];
    33             if(--in[to]==0) ak_47.push(to),deep[to]=deep[pos]+1,num[deep[to] ]++;
    34         }
    35         ak_47.pop();
    36     }
    37     if(cnt<n){
    38         cout<<-1<<endl;
    39         return ;
    40     }
    41     __int64 sum=0,tmp=888;
    42     for(int i=1;i<=n;i++)
    43         if(num[i]) sum+=num[i]*tmp,tmp++;
    44     cout<<sum<<endl;
    45 }
    46 int main(){
    47     while(cin>>n>>m){
    48         init();
    49         for(int i=1;i<=m;i++){
    50             int a,b;
    51             scanf("%d%d",&a,&b);
    52             in[a]++;
    53             p[b].pb(a);
    54         }
    55         tuopu();
    56     }
    57 }
  • 相关阅读:
    c/c++面试45-50之字符串
    c/c++面试39-44之内存动态分配
    使用spring配合Junit进行单元测试的总结
    使用springBoot进行快速开发
    配置项目使用weblogic的JNDI数据源
    转载-解决使用httpClient 4.3.x登陆 https时的证书报错问题
    SpringData JPA查询分页demo
    Lucene中的域选项
    代码片段,lucene基本操作(基于lucene4.10.2)
    配置maven使用nexus
  • 原文地址:https://www.cnblogs.com/ainixu1314/p/3896064.html
Copyright © 2011-2022 走看看