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2014多校7(1006)hdu4940(有上下界的最大流)

时间:2014-08-13 10:40:55      阅读:250      评论:0      收藏:0      [点我收藏+]

标签:acm   hdu   

Destroy Transportation system

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 114    Accepted Submission(s): 83


Problem Description
Tom is a commander, his task is destroying his enemy’s transportation system.

Let’s represent his enemy’s transportation system as a simple directed graph G with n nodes and m edges. Each node is a city and each directed edge is a directed road. Each edge from node u to node v is associated with two values D and B, D is the cost to destroy/remove such edge, B is the cost to build an undirected edge between u and v.

His enemy can deliver supplies from city u to city v if and only if there is a directed path from u to v. At first they can deliver supplies from any city to any other cities. So the graph is a strongly-connected graph.

He will choose a non-empty proper subset of cities, let’s denote this set as S. Let’s denote the complement set of S as T. He will command his soldiers to destroy all the edges (u, v) that u belongs to set S and v belongs to set T. 

To destroy an edge, he must pay the related cost D. The total cost he will pay is X. You can use this formula to calculate X:
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After that, all the edges from S to T are destroyed. In order to deliver huge number of supplies from S to T, his enemy will change all the remained directed edges (u, v) that u belongs to set T and v belongs to set S into undirected edges. (Surely, those edges exist because the original graph is strongly-connected)

To change an edge, they must remove the original directed edge at first, whose cost is D, then they have to build a new undirected edge, whose cost is B. The total cost they will pay is Y. You can use this formula to calculate Y:
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At last, if Y>=X, Tom will achieve his goal. But Tom is so lazy that he is unwilling to take a cup of time to choose a set S to make Y>=X, he hope to choose set S randomly! So he asks you if there is a set S, such that Y<X. If such set exists, he will feel unhappy, because he must choose set S carefully, otherwise he will become very happy.
 

Input
There are multiply test cases.

The first line contains an integer T(T<=200), indicates the number of cases. 

For each test case, the first line has two numbers n and m. 

Next m lines describe each edge. Each line has four numbers u, v, D, B. 
(2=<n<=200, 2=<m<=5000, 1=<u, v<=n, 0=<D, B<=100000)

The meaning of all characters are described above. It is guaranteed that the input graph is strongly-connected.
 

Output
For each case, output "Case #X: " first, X is the case number starting from 1.If such set doesn’t exist, print “happy”, else print “unhappy”.
 

Sample Input
2 3 3 1 2 2 2 2 3 2 2 3 1 2 2 3 3 1 2 10 2 2 3 2 2 3 1 2 2
 

Sample Output
Case #1: happy Case #2: unhappy
Hint
In first sample, for any set S, X=2, Y=4. In second sample. S= {1}, T= {2, 3}, X=10, Y=4.

题意:将图的点分成S和T集合,要求对任意的分法X<=Y始终成立

思路:其实可以将X看成流量的下界,将Y看成流量的上界,可以将问题转化为求一个无源无汇的有上下界的最大流

这个建图可以这样来:
  
           对于任意的节点i,令flow=∑流入的下界流量-∑流出的下界流量
    
          1.flow>0  说明节点i一定要流入flow的流量,则从附加源点向i连边,流量为flow
          
          2.flow<0  说明节点i一定要流出flow的流量,则从i向附加汇点连边,流量为flow

         然后跑最大流,跑完之后检查附加源点的每条边是否满流即可,如果满流则满足

2014多校7(1006)hdu4940(有上下界的最大流),布布扣,bubuko.com

2014多校7(1006)hdu4940(有上下界的最大流)

标签:acm   hdu   

原文地址:http://blog.csdn.net/cq_phqg/article/details/38524707

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