杭电2647--Reward（反向拓扑）-优快云博客

本文介绍了一种使用反向拓扑排序算法解决奖励分配问题的方法，通过构建有向图并进行拓扑排序，确保每位员工获得的奖励满足其要求且成本最低。案例分析和代码实现展示了该算法的应用。

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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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 }