zoukankan      html  css  js  c++  java
  • HDU-2647 Reward(拓扑排序)

    Reward

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 9799    Accepted Submission(s): 3131


    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
     
    Author
    dandelion
    #include<iostream>
    #include<cstring>
    #include<cstdio>
    #include<vector>
    #include<queue>
    
    using namespace std;
    const int N = 10000 + 15;
    int n, m, in[N], val[N];
    vector<int> edge[N];
    
    void Solve_question(){
        int ans = 0, cnt = 0;
        queue <int> Q;
    
        for(int i = 1; i <= n; i++)
            if(!in[i]) { Q.push(i); val[i] = 888; }
        while(!Q.empty()){
            int u = Q.front(); Q.pop();
            cnt++;
            for(int i = 0; i < (int)edge[u].size(); i++){
                int v = edge[u][i];
                if(--in[v] == 0){
                    Q.push(v);
                    val[v] = val[u] + 1;
                }
            }
        }
        if(cnt < n) puts("-1");
        else{
            for(int i = 1; i <= n; i++)
                ans += val[i];
            printf("%d
    ", ans);
        }
    }
    
    void Input_data(){
        for(int i = 1; i <= n; i++) edge[i].clear(), val[i] = in[i] = 0;
        int u, v;
        for(int i = 1; i <= m; i++){
            scanf("%d %d", &u, &v);
            in[u]++;
            edge[v].push_back(u);
        }
    }
    
    int main(){
        while(scanf("%d %d", &n, &m) == 2){
            Input_data();
            Solve_question();
        }
    }

    转载于:https://www.cnblogs.com/Pretty9/p/7413231.html

  • 相关阅读:
    java获得两个日期之间的所有月份
    Java设计模式之观察者模式
    SpringMVC项目配置
    Java设计模式之策略模式
    Tomcat源码
    线程池
    java内存模型
    JVM内存结构 JVM的类加载机制
    java虚拟机-垃圾回收算法
    并发容器-ConcurrentHashMap,CopyOnWriteArrayList
  • 原文地址:https://www.cnblogs.com/twodog/p/12139576.html
Copyright © 2011-2022 走看看