zoukankan      html  css  js  c++  java
  • HDU Problem 2647 Reward【拓扑排序】

    Reward

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 7391    Accepted Submission(s): 2317

    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
     

    思路:将图倒着存,如果找到一个节点的上一级,就用res数组保存这个节点上级所处的等级,最后做一个统计。

    #include <bits/stdc++.h>
    #define MAXN 10005
    using namespace std;
    int n, m, outdegree[MAXN], res[MAXN];
    int head[MAXN], num, que[MAXN];
    struct node{
        int from, to, next;
    }edge[MAXN*2];
    void add(int x, int y) {
        edge[num].from = x;
        edge[num].to = y;
        edge[num].next = head[x];
        head[x] = num++;
    }
    void init() {
        memset(head, -1, sizeof(head));
        memset(outdegree, 0, sizeof(outdegree));
        memset(res, 0, sizeof(res));
        num = 0;
    }
    void solved() {
        int iq = 0, ans = 0;
        for (int i = 1; i <= n; i++) {
            if (outdegree[i] == 0) {
                que[iq++] = i;
            }
        }
        for (int i = 0; i < iq; i++) {
            for (int k = head[que[i]]; k != -1; k = edge[k].next) {
                outdegree[edge[k].to]--;
                if (outdegree[edge[k].to] == 0) {
                    que[iq++] = edge[k].to;
                    //记录该节点所在的等级
                    res[edge[k].to] = res[edge[k].from] + 1;
    
                }
            }
        }
        for (int i = 1; i <= n; i++) {
            ans += res[i];
            //如果存在环
            if (outdegree[i] > 0) {
                printf("-1
    "); return ;
            }
        }
        printf("%d
    ", ans + 888*n);
    }
    int main() {
        int a, b;
        while (scanf("%d%d", &n, &m) != EOF) {
            init();
            for (int i = 0; i < m; i++) {
                scanf("%d%d", &a, &b);
                add(b, a); outdegree[a]++;
            }
            solved();
        }
        return 0;
    }
    
    


  • 相关阅读:
    jQuery——通过Ajax发送数据
    Python爬虫入门教程 71-100 续上篇,python爬虫爬取B站视频
    实战演练:PostgreSQL在线扩容
    直播丨Oracle比特币勒索&数据库大咖讲坛
    使用seaborn绘制强化学习中的图片
    nginx stream模块
    工具用的好下班走的早
    10年大数据平台经验,总结出这份数据建设干货(内含多张架构图)
    nginx 配置4层转发
    详解pytorch中的max方法
  • 原文地址:https://www.cnblogs.com/cniwoq/p/6770858.html
Copyright © 2011-2022 走看看