HDU 2647 Reward（拓扑排序）

Posted 2020-10-25 leonard-

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2647

题目：

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

 

题解：题目中有分层，用一个cost数组存一下每个点的花费，注意更新花费的时候是以最大值，因为可能这个在当前层的花费+1还没有其他层花费高，但是要满足所有层，就需要最大花费。

 1 #include <queue>
 2 #include <cstdio>
 3 #include <vector>
 4 #include <cstring>
 5 using namespace std;
 6 
 7 const int N=1e4+10;
 8 int cost[N],in[N];
 9 vector <int> E[N];
10 int n,m;
11 
12 bool toposort(){
13     int cnt=0;
14     queue <int> Q;
15     for(int i=1;i<=n;i++) if(!in[i]) Q.push(i);
16     while(!Q.empty()){
17         int u=Q.front();Q.pop();
18         cnt++;
19         for(int i=0;i<E[u].size();i++){
20             int v=E[u][i];
21             in[v]--;
22             cost[v]=max(cost[v],cost[u]+1);
23             if(in[v]==0) Q.push(v);
24         }
25     }
26     if(cnt==n) return true;
27     else return false;
28 }
29 
30 int main(){
31     int a,b;
32     while(scanf("%d %d",&n,&m)!=EOF){
33         for(int i=0;i<N;i++) E[i].clear();
34         memset(in,0,sizeof(in));
35         memset(cost,0,sizeof(cost));
36         for(int i=1;i<=m;i++){
37             scanf("%d %d",&a,&b);
38             E[b].push_back(a);
39             in[a]++;
40         }
41         if(!toposort()) printf("-1\n");
42         else{
43             int sum=0;
44             for(int i=1;i<=n;i++) sum+=cost[i];
45             printf("%d\n",sum+888*n);
46         }
47     }
48     return 0;
49 }