【杭电oj】2647 - Reward（拓扑排序）（含测试数据）

Reward

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6725    Accepted Submission(s): 2081

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

刚开始还以为是带权并查集，迅速的敲完，然后就WA了。

后来测试了一组这样的数据： 4 4 1 2 2 3 1 4 3 4

发现出问题了，后来才开始学习拓扑排序。

这道题要注意图应该是有向图，从谁出发到谁应该处理好，弄不好就懵了。

要是找不出bug的情况，不如试试以下类型的数据：

①普通题目中的数据（废话）

②有独立点的数据：3 1 1 2  ————  2665

③有环的数据（一种是无解的环，一种是有解的环）：3 3 1 2 2 3 3 1 ———— -1 

4 4 1 2 2 3 1 4 3 4 ———— 3558

代码如下：

#include <cstdio>
#include <queue>
#include <algorithm>
#include <cstring>
using namespace std;
struct node
{
	int from,to;		//表示方向 
}data[20022];

int n,m;
long long ans;
//int cost[10022];		//该点花费的钱数（为了计算下一个入度为0的节点） (好像这个多余了)
int used[10022];		//是否遍历过 
int in[10022];
int ach = 0;		//已完成的节点数 
bool flag;

void init()
{
	for (int i = 1 ; i <= n ; i++)
	{
		in[i] = 0;
		used[i] = 0;
	}
	ans = 0;
	flag = true;
}

void topo()
{
	queue<int>q;
	ach = 0;		//已完成的节点数 
	int s = 0;		//花费，每进行一轮花费加1（很朴素的方法，但是有效）
	while (1)
	{
		int k = 0;
		for (int i = 1 ; i <= n ; i++)
		{
			if (!used[i] && in[i] == 0)
			{
				q.push(i);
				k = 1;
				ans += s;
				ach++;
				used[i] = 1;
			}
		}
		if (k == 0)
			break;		//说明已经没有入度为0的点了
		while (!q.empty())
		{
			int tt = q.front();
			q.pop();
			for (int i = 1 ; i <= m ; i++)
			{
				if (data[i].from == tt && used[data[i].to] == 0)
					in[data[i].to]--;		//把边减掉 
			}
		}
		s++;
	}
}

int main()
{
	while (~scanf ("%d %d",&n,&m))
	{
		init();		//代码太乱，初始化忘了 = = 
		for (int i = 1 ; i <= m ; i++)
		{
			int t1,t2;
			scanf ("%d %d",&t1,&t2);		//这里注意需要反向，应该是t2指向t1
			data[i].from = t2;
			data[i].to = t1;
			in[t1]++;		//表示入度 
		}
		topo();
		if (ach != n)
			printf ("-1\n");
		else
		{
			ans += n * 888;
			printf ("%lld\n",ans);
		}
	}
	return 0;
}