HDU 2647 Reward （拓扑排序）

Reward

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7319    Accepted Submission(s): 2297

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 11 22 21 22 1

 

Sample Output

1777-1

 

Author

dandelion

 

Source

曾是惊鸿照影来

 

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思路：数据较大要用邻接表存储，对于相同等级的先用数组储存，

<pre name="code" class="cpp">#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int head[11000];int indegree[11000];int vis[11000];int num;struct node{int to,next;}edge[110000];void init(){memset(head,-1,sizeof(head));num=0;memset(indegree,0,sizeof(indegree));} void add(int a,int b){edge[num].to=b;edge[num].next=head[a];head[a]=num++;indegree[b]++;}int top(int n){int i,j,k,l,c,sum,num,t;int temp[11000];c=0;sum=0,j=0;num=888;memset(vis,0,sizeof(vis));while(1){memset(temp,0,sizeof(temp));k=0;for(i=1;i<=n;i++){if(!vis[i]&&indegree[i]==0)    {    vis[i]=1;    c++;    sum+=num;    temp[k++]=i;//把前驱数量为0的都放入数组         }}//printf("num1=%d\n",num);//printf("sum=%d\n",sum);    if(c>=n)    return sum;    if(!k)//存在环     return -1;    for(i=0;i<k;i++)//遍历所有前驱为0的结点     {    t=temp[i];    for(j=head[t];j!=-1;j=edge[j].next)    {    int tmp=edge[j].to;    indegree[tmp]--;}}num++;//printf("num=%d\n",num);} }int main(){int n,m,i,j,a,b;while(scanf("%d%d",&n,&m)!=EOF){init();for(i=0;i<m;i++){scanf("%d%d",&a,&b);add(b,a);//注意！！！！ }int ans=top(n);printf("%d\n",ans);}return 0;}