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
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <cassert>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#pragma comment(linker, "/stck:1024000000,1024000000")
#define lowbit(x) (x&(-x))
#define max(x,y) (x>=y?x:y)
#define min(x,y) (x<=y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.1415926535897932384626433832
#define ios() ios::sync_with_stdio(true)
#define INF 0x3f3f3f3f
#define mem(a) ((a,0,sizeof(a)))
typedef long long ll;
vector<int>v[];
int vis[],degree[];
int n,m,x,y,cnt,flag;
struct node
{
int x;
int y;
bool operator<(const node &a) const
{
return a.y<y;
}
}ans,pos;
void toposort()
{
priority_queue<node>q;
for(int i=;i<=n;i++)
{
if(!degree[i])
{
ans.x=i;
ans.y=;
vis[i]=;
q.push(ans);
}
}
while(!q.empty())
{
pos=q.top();
q.pop();
for(int i=;i<v[pos.x].size();i++)
{
degree[v[pos.x][i]]--;
if(!degree[v[pos.x][i]])
{
ans.x=v[pos.x][i];
ans.y=pos.y+;
q.push(ans);
vis[v[pos.x][i]]=pos.y+;
}
}
}
cnt=flag=;
for(int i=;i<=n;i++)
{
if(vis[i]==-) flag=;
else cnt+=+vis[i];
}
printf("%d\n",flag?-:cnt);
}
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
for(int i=;i<=n;i++)
v[i].clear();
memset(vis,-,sizeof(vis));
memset(degree,,sizeof(degree));
for(int i=;i<m;i++)
{
scanf("%d%d",&x,&y);
v[y].push_back(x);
degree[x]++;
}
toposort();
}
return ;
}

hdu 2647 Reward(拓扑排序+优先队列)的更多相关文章

  1. HDU.2647 Reward(拓扑排序 TopSort)

    HDU.2647 Reward(拓扑排序 TopSort) 题意分析 裸的拓扑排序 详解请移步 算法学习 拓扑排序(TopSort) 这道题有一点变化是要求计算最后的金钱数.最少金钱值是888,最少的 ...

  2. ACM: hdu 2647 Reward -拓扑排序

    hdu 2647 Reward Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Des ...

  3. HDU 2647 Reward (拓扑排序)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2647 题意是给你n点m条有向边,叶子点(出度为0)上的值为888,父亲点为888+1,依次计算... ...

  4. hdu 2647 Reward(拓扑排序+反图)

    题目链接:https://vjudge.net/contest/218427#problem/C 题目大意: 老板要给很多员工发奖金, 但是部分员工有个虚伪心态, 认为自己的奖金必须比某些人高才心理平 ...

  5. HDU 2647 逆向拓扑排序

    令每一个员工都有一个自己的等级level[i] , 员工等级越高,那么工资越高,为了使发的钱尽可能少,所以每一级只增加一单位的钱 输入a b表示a等级高于b,那么我们反向添加边,令b—>a那么i ...

  6. HDU 5638 Toposort 拓扑排序 优先队列

    Toposort 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5638 Description There is a directed acycli ...

  7. 题解报告:hdu 2647 Reward(拓扑排序)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2647 Problem Description Dandelion's uncle is a boss ...

  8. HDU 4857 拓扑排序 优先队列

    n个数,已经有大小关系,现给m个约束,规定a在b之前,剩下的数要尽可能往前移.输出序列 大小关系显然使用拓扑结构,关键在于n个数本身就有大小关系,那么考虑反向建图,优先选择值最大的入度为零的点,这样得 ...

  9. 2016"百度之星" - 初赛(Astar Round2A)HDU 5695 拓扑排序+优先队列

    Gym Class Time Limit: 6000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total S ...

随机推荐

  1. Install Rails on ubuntu 12.04 LTS

    There are basically there ways to install Rails development environment on your ubuntu linux system, ...

  2. android之handler机制深入解析

    一.android中需要另开线程处理耗时.网络的任务,但是有必须要在UI线程中修改组件.这样做是为了: ①只能在UI线程中修改组件,避免了多线程造成组件显示混乱 ②不使用加锁策略是为了提高性能,因为a ...

  3. RocketMQ学习笔记(7)----RocketMQ的整体架构

    1. RocketMQ主要的9个模块,如图: 2. 模块介绍 1. rocketmq-common:通用的常量枚举,基类方法或者数据结构,按描述的目标来分包,通俗易懂.报名有admin,consume ...

  4. for 的相关用法

    forEach() <!DOCTYPE html> <html> <head> <meta charset="utf-8"> < ...

  5. linux查看系统cpu信息

    # 查看物理CPU个数 cat /proc/cpuinfo| grep "physical id"| sort| uniq| wc -l # 查看每个物理CPU中core的个数(即 ...

  6. 队列(Queue)-c实现

    相对而言,队列是比较简单的. 代码还有些warning,我改不动,要找gz帮忙. #include <stdio.h> typedef struct node { int data; st ...

  7. 使用vue实现简单键盘,支持移动端和pc端

    常看到各种app应用中使用自定义的键盘,本例子中使用vue2实现个简单的键盘,支持在移动端和PC端使用,欢迎点赞,h5 ios输入框与键盘 兼容性优化 实现效果: Keyboard.vue <t ...

  8. 2018秋寒假作业4——PTA编辑总结1

    #include<stdio.h> #include<math.h> int main(void) { int n,i,j,p,m,ge,N,k; char op; ){ sc ...

  9. thymeleaf 常用标签

    1.th:field th:field="*{user.sex}" 此标签会自动填充数据,比如用户的性别 user.sex 如果不为空,则会自动勾选上  2.th:each=&qu ...

  10. 【Henu ACM Round#24 B】Gargari and Bishops

    [链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 如果写过n皇后问题. 肯定都知道 某个点(i,j)和它在同一条对角线上的点分别是i+j的值和i-j的值相同的点. 然后会发现选择的两 ...