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.
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.
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
Recommend
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 ", total); 45 } 46 else 47 printf("-1 "); 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 }