hdu2647——Reward

Reward

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4306    Accepted Submission(s): 1318

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
   
   
   
   

 

Author

dandelion

 

Source

曾是惊鸿照影来

 

Recommend

一开始我以为是差分约束，然后写了下发现超时，无奈去看了下发现是topo排序，然后改正后就AC了

#include <map>  
#include <set>  
#include <list>  
#include <stack>  
#include <vector>  
#include <queue>  
#include <cmath>  
#include <cstdio>  
#include <cstring>  
#include <iostream>  
#include <algorithm>  

using namespace std;

const int N = 10010;
const int M = 20010;

int tot, n, m;
int head[N];
int num[N];
int in_degree[N];

struct node
{
	int next;
	int to;
}edge[M];

void addedge(int from, int to)
{
	edge[tot].to = to;
	edge[tot].next = head[from];
	head[from] = tot++;
}

bool topo_sort()
{
	queue <pair<int, int> > du;
	memset (num, 0, sizeof(num) );
	int cnt = 0;
	while ( !du.empty() )
	{
		du.pop();
	}
	for (int i = 1; i <= n; ++i)
	{
		if (!in_degree[i])
		{
			du.push( make_pair(0, i) );
			num[i] = 0;
		} 
	}
	while ( !du.empty() )
	{
		pair<int, int> tmp = du.front();
		du.pop();
		cnt++;
		int u = tmp.second;
		int res = tmp.first;
		for (int i = head[u]; ~i; i = edge[i].next)
		{
			in_degree[edge[i].to]--;
			if ( !in_degree[edge[i].to] )
			{
				num[edge[i].to] = res + 1;
				du.push( make_pair(res + 1, edge[i].to) );
			}
		}
	}
	return cnt == n;
}

int main()
{
	while (~scanf("%d%d", &n, &m))
	{
		int u, v;
		memset (head, -1, sizeof(head) );
		memset (in_degree, 0, sizeof(in_degree) );
		tot = 0;
		for (int i = 0; i < m; ++i)
		{
			scanf("%d%d", &u, &v);
			addedge(v, u);
			in_degree[u]++;
		}
		bool flag = topo_sort();
		if (!flag)
		{
			printf("-1\n");
			continue;
		}
		int ans = 0;
		for (int i = 1; i <= n; ++i)
		{
			ans += num[i];
		}
		ans += 888 * n;
		printf("%d\n", ans);
	}
	return 0;
}