zoukankan      html  css  js  c++  java
  • POJ

    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 profitVi 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 cityi, Vi (0 ≤ |Vi| ≤ 20000)
    The next m lines each contain two integers x, y indicating that there is a road leads from cityx 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

    题意:一个人去找工作遇到了一道面试题。面试官要求给出一些城市和城市之间的道路,每到达一个城市。可能会赚一些钱,可是也可能会有损失。

    终于面试者的所得会决定他能否得到这份工作。显而易见,越多越好。

    思路:由于是有向无环图(DAG)并且事实上求的是从一个0入度到0出度的路径,所以我们能够用topsort来处理,再加上简单的DP 即可了

    #include <iostream>
    #include <cstring>
    #include <cstdio>
    #include <algorithm>
    #include <queue>
    #include <vector>
    using namespace std;
    const int maxn = 100005;
    const int inf = 0x3f3f3f3f;
    
    struct Node {
    	int v, next;
    }node[maxn*20];
    int n, m, cnt;
    int profit[maxn];
    int ind[maxn], out[maxn], dp[maxn], adj[maxn];
    
    void topsort() {
    	queue<int> q;
    	for (int i = 1; i <= n; i++) 
    		if (ind[i] == 0) {
    			q.push(i);
    			dp[i] = profit[i];
    		}
    	while (!q.empty()) {
    		int cur = q.front();
    		q.pop();
    		for (int i = adj[cur]; i != -1; i = node[i].next) {
    			int v = node[i].v;
    			if (dp[v] < dp[cur]+profit[v])
    				dp[v] = dp[cur]+profit[v];
    			if (--ind[v] == 0)
    				q.push(v);
    		}
    	}
    }
    
    int main() {
    	while (scanf("%d%d", &n, &m) != EOF) {
    		cnt = 0;
    		memset(adj, -1, sizeof(adj));
    		memset(ind, 0, sizeof(ind));
    		memset(out, 0, sizeof(out));
    		for (int i = 1; i <= n; i++) {
    			dp[i] = -inf;
    			scanf("%d", &profit[i]);
    		}
    		for (int i = 0; i < m; i++) {
    			int a, b;
    			scanf("%d%d", &a, &b);
    			out[a]++;
    			ind[b]++;
    			node[cnt].v = b;
    			node[cnt].next = adj[a];
    			adj[a] = cnt++;
    		}
    		topsort();
    		int ans = -inf;
    		for (int i = 1; i <= n; i++) 
    			if (out[i] == 0 && dp[i] > ans)
    				ans = dp[i];
    		printf("%d
    ", ans);
    	}
    	return 0;
    }


  • 相关阅读:
    程序员的自我修养
    c++中的const 限定修饰符
    基于.net开发平台项目案例集锦
    中国期货公司排行及相关上市公司
    备份一些好的书籍名字
    商业银行房贷业务节后骤然下降
    散户炒股七大绝招 巨额获利风险小 (网摘)
    上海2月住宅供应剧减七成 房企捂盘保价
    2009年中国各省人均GDP排名(鄂尔多斯人均GDP将很有可能超过两万美元,全国第一)
    (载自MSN )个人炒汇多年来的一些心得
  • 原文地址:https://www.cnblogs.com/bhlsheji/p/5058116.html
Copyright © 2011-2022 走看看