Test for Job
Time Limit: 5000MS   Memory Limit: 65536K
Total Submissions: 13457   Accepted: 3100

Description

Mr.Dog was fired by his company. In order to support his family, he must find a new job as soon as possible. Nowadays, It's hard to have a job, since there are swelling numbers of the unemployed. So some companies often use hard tests for their recruitment.

The test is like this: starting from a source-city, you may pass through some directed roads to reach another city. Each time you reach a city, you can earn some profit or pay some fee, Let this process continue until you reach a target-city. The boss will compute the expense you spent for your trip and the profit you have just obtained. Finally, he will decide whether you can be hired.

In order to get the job, Mr.Dog managed to obtain the knowledge of the net profit Vi of all cities he may reach (a negative Vi indicates that money is spent rather than gained) and the connection between cities. A city with no roads leading to it is a source-city and a city with no roads leading to other cities is a target-city. The mission of Mr.Dog is to start from a source-city and choose a route leading to a target-city through which he can get the maximum profit.

Input

The input file includes several test cases. 
The first line of each test case contains 2 integers n and m(1
≤ n ≤ 100000, 0 ≤ m ≤ 1000000) indicating the number of cities
and roads. 
The next n lines each contain a single integer. The ith line
describes the net profit of the city iVi (0 ≤ |Vi|
≤ 20000) 
The next m lines each contain two integers xy indicating
that there is a road leads from city x to city y. It is
guaranteed that each road appears exactly once, and there is no way to
return to a previous city. 
 

Output

The output file contains one line for each test cases, in which contains an
integer indicating the maximum profit Dog is able to obtain (or the minimum
expenditure to spend)

Sample Input

6 5
1
2
2
3
3
4
1 2
1 3
2 4
3 4
5 6

Sample Output

7

Hint

Source

【题意】

给n个点,m条单向边,每个点有点权,没有边权,然后遍历

一条路径:以任意一个入度为0的点为起点,沿着单向边走,走到任意一个出度为0的点为终点,路径长度为经过个点的点权之和。

求最长路径。

【分析】

一开始想的是拓扑排序,分离出起点和终点,然后dijkstra跑一遍。结果,无论算时间复杂度,还是实际交付评测,都会T。

然后,再读题目,发现每个点的后继都唯一,然后,满足无后效性,猜想DP,递推代码不好打,直接从每个起点记忆话搜索,跑最长路。

保证时间复杂度:O(m)

【代码】
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#define m(s) memset(s,0,sizeof s)
using namespace std;
const int N=1e5+5;
int n,m,val[N],du[N],f[N];
vector<int>e[N];
inline void Clear(){
m(du);m(f);
for(int i=1;i<=n;i++) e[i].clear();
}
inline void Init(){
for(int i=1;i<=n;i++) scanf("%d",val+i);
for(int i=1,x,y;i<=m;i++) scanf("%d%d",&x,&y),du[y]++,e[x].push_back(y);
}
int dfs(int x){
int &now=f[x];
if(now) return now;
now=-2e9;
if(!e[x].size()) return now=val[x];
for(int i=0;i<e[x].size();i++){
int v=e[x][i];
now=max(now,val[x]+dfs(v));
}
return now;
}
inline void Solve(){
int ans=-2e9;
for(int i=1;i<=n;i++){
if(!du[i]){
ans=max(ans,dfs(i));
}
}
printf("%d\n",ans);
}
int main(){
while(scanf("%d%d",&n,&m)==2){
Clear();
Init();
Solve();
}
return 0;
}

 

POJ 3249 Test for Job的更多相关文章

  1. POJ 3249 Test for Job (拓扑排序+DP)

    POJ 3249 Test for Job (拓扑排序+DP) <题目链接> 题目大意: 给定一个有向图(图不一定连通),每个点都有点权(可能为负),让你求出从源点走向汇点的路径上的最大点 ...

  2. poj 3249 Test for Job (记忆化深搜)

    http://poj.org/problem?id=3249 Test for Job Time Limit: 5000MS   Memory Limit: 65536K Total Submissi ...

  3. poj 3249(bfs+dp或者记忆化搜索)

    题目链接:http://poj.org/problem?id=3249 思路:dp[i]表示到点i的最大收益,初始化为-inf,然后从入度为0点开始bfs就可以了,一开始一直TLE,然后优化了好久才4 ...

  4. POJ 3249 Test for Job (dfs + dp)

    题目链接:http://poj.org/problem?id=3249 题意: 给你一个DAG图,问你入度为0的点到出度为0的点的最长路是多少 思路: 记忆化搜索,注意v[i]可以是负的,所以初始值要 ...

  5. poj 3249 Test for Job (DAG最长路 记忆化搜索解决)

    Test for Job Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 8990   Accepted: 2004 Desc ...

  6. Test for Job (poj 3249 记忆化搜索)

    Language: Default Test for Job Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 9733   A ...

  7. POJ 3249 拓扑排序+DP

    貌似是道水题.TLE了几次.把所有的输入输出改成scanf 和 printf ,有吧队列改成了数组模拟.然后就AC 了.2333333.... Description: MR.DOG 在找工作的过程中 ...

  8. poj 3249 拓扑排序 and 动态规划

    思路:我们首先来一遍拓扑排序,将点按先后顺序排列于一维数组中,然后扫描一遍数组,将每个点的出边所连接的点进行更新,即可得到最优解. #include<iostream> #include& ...

  9. POJ - 3249 Test for Job (DAG+topsort)

    Description Mr.Dog was fired by his company. In order to support his family, he must find a new job ...

随机推荐

  1. Kubernetes命名空间

    本文环境为Kubernetes V1.11,操作系统版本为 CentOs 7.3,Kubernetes集群安装可以参考 kubeadm安装kubernetes V1.11.1 集群 1. 什么是Nam ...

  2. 修改文件夹的protection level之后,哪个job会来执行re-stripe的操作呢?

    有下面的一些job可能参与其中的,他们的描述如下: AutoBalance,AutoBalanceLin - Balances free space in the cluster. The goal ...

  3. centos7 使用rsync 实现文件同步

    一.服务端(192.168.8.81): 安装软件: yum -y install rsync 创建需同步的目录: mkdir -p /home/root/rsync 编辑配置文件:vim /etc/ ...

  4. Java+大数据开发——Hadoop集群环境搭建(一)

    1集群简介 HADOOP集群具体来说包含两个集群:HDFS集群和YARN集群,两者逻辑上分离,但物理上常在一起 HDFS集群: 负责海量数据的存储,集群中的角色主要有 NameNode / DataN ...

  5. 【Spark】Spark-架构

    Spark-架构 Spark Master at spark://node-01:7077 spark clustermanager_百度搜索 看了之后不再迷糊-Spark多种运行模式 - 简书 Sp ...

  6. Revit中如何给不同构件着色

    在Revit构件密集,默认的显示模式难以区分不同构件的区别,比如建筑立面有很多不同的机电管道,风管.水管,电缆桥架等,可一个给不同的机电管线添加不同的颜色,以示其区别,如下图所示,完成着色后,各种不同 ...

  7. 【T06】记住TCP是一种流协议

    1.TCP是一种流协议(stream protocol),这意味着数据是以字节流的形式发给接收者的,没有固定的报文和报文边界的概念. 接收端读取tcp数据,无法预知在这一次读操作中会返回多少个字节. ...

  8. PHP中日志相关处理

    内置函数: 1.error_log() ,第三个参数不能是绝对路径,必须是相对路径.写入文件: error_log("warn:\nthis is a warn!\n",3,&qu ...

  9. llvm Array Bounds Check Elimination

    http://www.knosof.co.uk/vulnerabilities/arraybnd.html http://www.cs.utsa.edu/dmz/techrep/2010/CS-TR- ...

  10. Series转化为DataFrame数据

    out=groupby_sum.ix[:'to_uid','sum(diamonds)']使用ix在提取数据的时候,out的数据类型通常为<class 'pandas.core.series.S ...