杭电2647--Reward（反向拓扑）

最新推荐文章于 2025-09-09 09:10:31 发布

转载最新推荐文章于 2025-09-09 09:10:31 发布 · 73 阅读

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原文链接：http://www.cnblogs.com/soTired/p/4743928.html

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#java

解决一个工厂老板如何在满足员工提出的相对金额要求下，使用最少的资金进行奖金分配的问题。采用反向拓扑排序的方法，并确保每位员工的奖金不低于888。

Reward

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5863 Accepted Submission(s): 1797

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

Author

dandelion

Source

曾是惊鸿照影来

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yifenfei | We have carefully selected several similar problems for you: 1811 1142 2729 1548 3339

WA了好多次，题目意思是：如果一个人在一个人之后领奖金，他会比前一个人领的多。所以说这个人之后可能会同时存在两个人领的钱数一样多。反向拓扑。数据较多，只能用邻接表。

 1 #include <queue>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <iostream>
 5 using namespace std;
 6 int indegree[10010], mon[10010]; 
 7 int n, m; 
 8 struct Edge
 9 {
10     int from, to, next;
11 } edge[20020];
12 int head[10010], cnt; 
13 void Add(int a, int b)
14 {
15     Edge E = {a, b, head[a]}; 
16     edge[cnt] = E;
17     head[a] = cnt++;
18 } 
19 int sum;
20 void Tsort()
21 {
22     sum = 0;
23     queue<int> q;
24     for(int i = 1; i <= n; i++){
25         if(!indegree[i])
26             q.push(i);
27     mon[i] = 888;
28     }
29     while(!q.empty())
30     {
31         int u = q.front();
32         q.pop(); indegree[u]--;  sum++;
33         for(int i = head[u]; i != -1; i = edge[i].next)
34         {
35             if(!--indegree[edge[i].to])
36                 mon[edge[i].to] = mon[u] + 1, q.push(edge[i].to); 
37         }
38     }
39     int total = 0;
40     if(sum == n)
41     {
42         for(int i = 1; i <= n; i++)
43             total += mon[i];
44         printf("%d\n", total);
45     }
46     else
47         printf("-1\n"); 
48 }
49 int main()
50 {
51     while(~scanf("%d %d", &n, &m))
52     {
53         cnt = 0;
54         memset(head, -1, sizeof(head));
55         memset(indegree, 0, sizeof(indegree));
56         for(int i = 1; i <= m; i++)
57         {
58             int a, b;
59             scanf("%d %d", &a, &b);
60             indegree[a]++;
61             Add(b, a);    
62         } 
63         Tsort();
64     } 
65     return 0;    
66 }