Test for Job
Time Limit: 5000MS   Memory Limit: 65536K
Total Submissions: 10137   Accepted: 2348

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

题目链接:点击打开链接

给出n个点的权值和m条路径, 问一条路的最大权值是多少.

起点入度为0, 对全部入度为0的点赋值到dp数组, 进行拓扑排序, 最后遍历出度为0的点, 求得最大值.

AC代码:

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
#include "queue"
#include "stack"
#include "cmath"
#include "utility"
#include "map"
#include "set"
#include "vector"
#include "list"
#include "string"
using namespace std;
typedef long long ll;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int MAXN = 1e5 + 5;
int n, m, num;
int cost[MAXN], in[MAXN], out[MAXN], head[MAXN], dp[MAXN];
bool vis[MAXN];
struct node
{
/* data */
int fr, to, nxt;
}e[MAXN * 10];
void add(int x, int y)
{
e[num].fr = x;
e[num].to = y;
e[num].nxt = head[x];
head[x] = num++;
}
void toposort()
{
int cnt = 1;
while(cnt < n) {
for(int i = 1; i <= n; ++i)
if(in[i] == 0 && !vis[i]) {
vis[i] = true;
cnt++;
for(int j = head[i]; j != -1; j = e[j].nxt) {
int x = e[j].to;
in[x]--;
if(dp[i] + cost[x] > dp[x]) dp[x] = dp[i] + cost[x];
}
}
}
}
int main(int argc, char const *argv[])
{
while(scanf("%d%d", &n, &m) != EOF) {
memset(in, 0, sizeof(in));
memset(out, 0, sizeof(out));
memset(head, -1, sizeof(head));
memset(vis, false, sizeof(vis));
num = 1;
for(int i = 1; i <= n; ++i)
scanf("%d", &cost[i]);
for(int i = 1; i <= m; ++i) {
int x, y;
scanf("%d%d", &x, &y);
add(x, y);
in[y]++;
out[x]++;
}
for(int i = 1; i <= n; ++i)
if(in[i] == 0) dp[i] = cost[i];
else dp[i] = -INF;
toposort();
int ans = -INF;
for(int i = 1; i <= n; ++i)
if(out[i] == 0 && dp[i] > ans) ans = dp[i];
printf("%d\n", ans);
}
return 0;
}

POJ3249 Test for Job(拓扑排序+dp)的更多相关文章

  1. 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 ...

  2. [NOIP2017]逛公园 最短路+拓扑排序+dp

    题目描述 给出一张 $n$ 个点 $m$ 条边的有向图,边权为非负整数.求满足路径长度小于等于 $1$ 到 $n$ 最短路 $+k$ 的 $1$ 到 $n$ 的路径条数模 $p$ ,如果有无数条则输出 ...

  3. 洛谷P3244 落忆枫音 [HNOI2015] 拓扑排序+dp

    正解:拓扑排序+dp 解题报告: 传送门 我好暴躁昂,,,怎么感觉HNOI每年总有那么几道题题面巨长啊,,,语文不好真是太心痛辣QAQ 所以还是要简述一下题意,,,就是说,本来是有一个DAG,然后后来 ...

  4. 【BZOJ-1194】潘多拉的盒子 拓扑排序 + DP

    1194: [HNOI2006]潘多拉的盒子 Time Limit: 10 Sec  Memory Limit: 162 MBSubmit: 456  Solved: 215[Submit][Stat ...

  5. 【BZOJ5109】[CodePlus 2017]大吉大利,晚上吃鸡! 最短路+拓扑排序+DP

    [BZOJ5109][CodePlus 2017]大吉大利,晚上吃鸡! Description 最近<绝地求生:大逃杀>风靡全球,皮皮和毛毛也迷上了这款游戏,他们经常组队玩这款游戏.在游戏 ...

  6. bzoj1093[ZJOI2007]最大半连通子图(tarjan+拓扑排序+dp)

    Description 一个有向图G=(V,E)称为半连通的(Semi-Connected),如果满足:?u,v∈V,满足u→v或v→u,即对于图中任意两点u,v,存在一条u到v的有向路径或者从v到u ...

  7. 【bzoj4011】[HNOI2015]落忆枫音 容斥原理+拓扑排序+dp

    题目描述 给你一张 $n$ 个点 $m$ 条边的DAG,$1$ 号节点没有入边.再向这个DAG中加入边 $x\to y$ ,求形成的新图中以 $1$ 为根的外向树形图数目模 $10^9+7$ . 输入 ...

  8. 【bzoj1093】[ZJOI2007]最大半连通子图 Tarjan+拓扑排序+dp

    题目描述 一个有向图G=(V,E)称为半连通的(Semi-Connected),如果满足:对于u,v∈V,满足u→v或v→u,即对于图中任意两点u,v,存在一条u到v的有向路径或者从v到u的有向路径. ...

  9. 【bzoj4562】[Haoi2016]食物链 拓扑排序+dp

    原文地址:http://www.cnblogs.com/GXZlegend/p/6832118.html 题目描述 如图所示为某生态系统的食物网示意图,据图回答第1小题 现在给你n个物种和m条能量流动 ...

随机推荐

  1. javascript中的分号【;】

    以前一直以为,在编写js代码的时候,如果在代码后面不添加分号,JavaScript会自动填补分号.最近看了权威指南,才突然发现一直理解有误,而且关于分号的使用,还有很多需要注意的地方. 1.分号的省略 ...

  2. flex和layout移动端布局

    1.九宫格 样式为: ul{ display: flex; flex-wrap: wrap;//超出换行 } li{ width: 33%; height: 60px; display: flex; ...

  3. ML一些简单的资源

    参考文献及推荐阅读 维基百科,http://en.wikipedia.org/wiki/K-nearest_neighbor_algorithm: 机器学习中的相似性度量,http://www.cnb ...

  4. JS 封装一个求n~m的求和函数

    var a = 0;    cc(2,10);    function cc(n,m){        for(var i =n;i<(m+1);i++){            a = a + ...

  5. Brain Network (hard) CodeForces - 690C 简单倍增 + 一些有趣的推导

    Code: #include<cstdio> #include<cstring> #include<algorithm> using namespace std; ...

  6. python文件操作IO

    模式 描述 r 以只读方式打开文件.文件的指针将会放在文件的开头.这是默认模式. rb 以二进制格式打开一个文件用于只读.文件指针将会放在文件的开头.这是默认模式.一般用于非文本文件如图片等. r+ ...

  7. 用div布局,页面copyright部分始终居于

    <!DOCTYPE HTML><html><head><meta http-equiv="Content-Type" content=&q ...

  8. HDU 4704 Sum( 费马小定理 + 快速幂 )

    链接:传送门 题意:求 N 的拆分数 思路: 吐嘈:求一个数 N 的拆分方案数,但是这个拆分方案十分 cd ,例如:4 = 4 , 4 = 1 + 3 , 4 = 3 + 1 , 4 = 2 + 2 ...

  9. [SDOI2016]生成魔咒(后缀自动机)

    看一眼题.本质不同的字串数. 嘴角微微上扬. 每一次加一个数输出一个答案. 笑容渐渐消失. 等等,\(SAM\)好像也可以求本质不同的字串. 设当前字符串用\(x\)表示,每次插入完成后\(ans\) ...

  10. 1.1 Eclipse的安装

    下载地址:http://www.eclipse.org/downloads/packages/eclipse-ide-java-ee-developers/indigosr1 2.java jdk 的 ...