HDU 2647 Reward

Reward

Time Limit: 1000ms

Memory Limit: 32768KB

This problem will be judged on  HDU. Original ID: 2647
64-bit integer IO format: %I64d      Java class name: Main

 

 

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

Source

曾是惊鸿照影来

 

解题：拓扑排序。

 

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <cmath>

 5 #include <algorithm>

 6 #include <climits>

 7 #include <vector>

 8 #include <queue>

 9 #include <cstdlib>

10 #include <string>

11 #include <set>

12 #include <stack>

13 #define LL long long

14 #define pii pair<int,int>

15 #define INF 0x3f3f3f3f

16 using namespace std;

17 const int maxn = 10010;

18 int n,m,in[maxn],money[maxn];

19 vector<int>g[maxn];

20 queue<int>q;

21 int topSort() {

22     int ans = 0;

23     while(!q.empty()) q.pop();

24     for(int i = 1; i <= n; i++) {

25         if(!in[i]) {

26             q.push(i);

27             money[i] = 888;

28         }

29     }

30     int sum = n;

31     while(!q.empty()){

32         sum--;

33         int u = q.front();

34         q.pop();

35         for(int i = 0; i < g[u].size(); i++){

36             if(--in[g[u][i]] == 0){

37                 q.push(g[u][i]);

38                 money[g[u][i]] = money[u]+1;

39             }

40         }

41     }

42     if(sum > 0) return -1;

43     else{

44         for(int i = 1; i <= n; i++)

45             ans += money[i];

46         return ans;

47     }

48 }

49 int main() {

50     int i,u,v;

51     while(~scanf("%d %d",&n,&m)) {

52         for(i = 1; i <= n; i++) {

53             g[i].clear();

54             in[i] = 0;

55             money[i] = 0;

56         }

57         for(i = 0; i < m; i++) {

58             scanf("%d %d",&v,&u);

59             g[u].push_back(v);

60             in[v]++;

61         }

62         printf("%d\n",topSort());

63     }

64     return 0;

65 }

View Code