zoukankan      html  css  js  c++  java
  • Test for Job 图上的动态规划(DAG)

    Test for Job
    Time Limit: 5000MS   Memory Limit: 65536K
    Total Submissions: 11399   Accepted: 2697

    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

     
    按照拓扑序列进行DP
    #include <iostream>
    #include <string>
    #include <cstdio>
    #include <vector>
    #include <algorithm>
    #include <deque>
    #include <map>
    #include <stack>
    #include <cstring>
    using namespace std;
    typedef long long LL;
    #define INF 0x3f3f3f3f
    #define MAXN 200000+9
    #define MAXM 2000000 + 9
    
    
    struct edge
    {
        LL next, v, cost;
    }E[MAXM];
    LL n, m;
    LL head[MAXN], tot;
    LL cnt[MAXN], topsort[MAXN];
    LL val[MAXN], dist[MAXN],out[MAXN];
    void init()
    {
        tot = 0;
        memset(head, -1, sizeof(head));
        memset(cnt, 0, sizeof(cnt));
        memset(out, 0, sizeof(out));
    }
    void addedge(LL f, LL t, LL d)
    {
        E[tot].v = t;
        E[tot].cost = d;
        E[tot].next = head[f];
        head[f] = tot++;
    }
    LL Topsort()
    {
        LL p = 0;
        for (LL i = 1; i <= n; i++)
        {
            if (cnt[i] == 0)
                dist[i] = val[i], topsort[p++] = i, cnt[i]--;
            else
                dist[i] = -INF;
        }
        for (LL i = 0; i < p; i++)
        {
            for (LL j = head[topsort[i]]; j != -1; j = E[j].next)
            {
                LL v = E[j].v;
                dist[v] = max(dist[v], dist[topsort[i]] + E[j].cost);
                if (--cnt[v] == 0)
                    topsort[p++] = v;
            }
        }
        LL ans = -INF;
        for (int i = 1; i <= n; i++)
            if (!out[i])
                ans = max(ans, dist[i]);
        return ans;
    
    }
    
    int main()
    {
        while (scanf("%lld%lld", &n, &m) != EOF)
        {
            init();
            for (int i = 1; i <= n; i++)
                scanf("%lld", &val[i]);
            LL f, t;
            for (int i = 1; i <= m; i++)
            {
                scanf("%lld%lld", &f, &t);
                addedge(f, t, val[t]);
                cnt[t]++;
                out[f]++;
            }
            printf("%lld
    ", Topsort());
        }
    }
  • 相关阅读:
    SQL常用语句(不定期更新)包含 日期格式,取列
    前端开发 ---浏览器自动刷新
    记录Redis使用中遇到的两个问题(原子性及数据完整性)
    Linux 输出重定向
    逆向手机内核,添加调试支持及绕过反调试
    阿里2014移动安全挑战赛第二题调试笔记
    在redhat6.4上编译z3求解器
    WCF学习笔记一(概述)
    DataReader反射泛型对象
    自定义配置节点(一)
  • 原文地址:https://www.cnblogs.com/joeylee97/p/7614667.html
Copyright © 2011-2022 走看看