Reward 2647 （拓扑排序+邻接表）

Reward

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5701    Accepted Submission(s): 1730

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
   
   
   
   
   
   
   
   

    
    
    
    
#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
using namespace std;
struct zz
{
	int to;
	int next;
}p[51000];
int in[51000];
int head[51000];
int num[51000];
void topo(int n)
{
	queue<int>q;
	int i,j,k;
	int ans=0;
	int sum=0;
	for(i=1;i<=n;i++)
	{
		if(in[i]==0)
		q.push(i);
	}
	while(!q.empty())
	{
		k=q.front();
		q.pop();
		ans++;
		for(j=head[k];j!=-1;j=p[j].next)
		{
			in[p[j].to]--;
			if(in[p[j].to]==0)
			{
				q.push(p[j].to);
				num[p[j].to]=num[k]+1;
			}
		}
	}
	for(i=1;i<=n;i++)
		sum+=num[i];
		if(ans!=n)
			printf("-1\n");
		else
			printf("%d\n",sum);
}
int main(){
	int n,m,i,j,a,b;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		memset(in,0,sizeof(in));
		memset(head,-1,sizeof(head));
		for(i=1;i<=n;i++)
			num[i]=888;
		for(i=0;i<m;i++)
		{
			scanf("%d%d",&a,&b);
			p[i].to=a;
			p[i].next=head[b];
			head[b]=i;
			in[a]++;
		}
		topo(n);
	}
	return 0;
}
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
    //大神的代码
   
   
   
   
   
   
   
   

    
    
    
    
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<queue>
using namespace std;
int n,m;
int indegree[10010];
struct Node{
	int to;
	int next;
}A[20010];
int head[10010];
int value[10010];
void INPUT()
{
	int i,j,a,b;
	memset(head,-1,sizeof(head));
	memset(indegree,0,sizeof(indegree));
	for(i=0;i<m;++i)
	{
		scanf("%d%d",&a,&b);
		A[i].to=a;
		A[i].next=head[b];
		head[b]=i;
		indegree[a]++;
	}
}
void Topo(){
	int i,j;
	queue<int>Q;
	for(i=1;i<=n;++i)
	{
		value[i]=888;
		if(indegree[i]==0)
		{
			Q.push(i);
		}
	}
	int top,money=0,ans=0;
	while(!Q.empty())
	{
		ans++;
		top=Q.front();Q.pop();
		indegree[top]=-1;
		money+=value[top];
		for(int k=head[top];k!=-1;k=A[k].next)
		{
			indegree[A[k].to]--;
			if(indegree[A[k].to]==0)
			{
				Q.push(A[k].to);
			}
			value[A[k].to]=value[top]+1;
		}
	}
	if(ans==n)
		printf("%d\n",money);
	else
		printf("-1\n");
}
int main()
{
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		INPUT();
		Topo();
	}
	return 0;
}