1459power network解题报告by解法或类型网络流

1、er Network解题题目来源：http解法或类型：网络流作者：/JudgeOnline/showproblem?problem_id=1459（No.1459）题目描述er NetworkTime Limit:2SMemory Limit:1000KTotal Submit:103 Accepted:40DescriptionAer network nected bysists of nodes (er sions,sumers and dispatchers) d win amounter transport lines. A node u may bes(u) = 0 ofer, ma

2、y produce an amount 0 = p(u) = pmax(u) ofer, maysume an amount 0 = c(u) = min(s(u),cmax(u) ofer, and may deliver anamount d(u)=s(u)+p(u)-c(u) ofer. The following restrictions apply: c(u)=0 foranyer sion, p(u)=0 for anysumer, and p(u)=c(u)=0 for any dispatcher.There is at most oneer transport line (u

3、,v) from a node u to a node vhe net; ittransports an amount 0 = l(u,v) = lmax(u,v) ofer delivered by u to v. Let=uc(u) be the um value ofer.sumedhe net. The problem is to compute theAn exle is in figure 1. The label x/y ofer sion u showst p(u)=x andpmax(u)=y. The label x/y ofsumer u showst c(u)=x an

4、d cmax(u)=y. The label x/yofer transport line (u,v) showst l(u,v)=x and lmax(u,v)=y. Theersumed is=6. Noticet there are othossible ses of the network but thevalue ofcannot exceed 6.InputThere are several data setshe input. Each data set encodes aer network. Iter sions), 0 =starts with fouregers: 0 =

5、 n = 100 (nodes), 0 = np = n (nc = n (sumers), and 0 = m = n2 (er transport lines). Follow m datatriplets (u,v)z, where u and v are node identifiers (starting from 0) and 0 = z = 1000is the value of lmax(u,v). Follow np doublets (u)z, where u is the identifier of aersion and 0 = z = 10000 is the val

6、ue of pmax(u). The data set ends with nc doublets(u)z, where u is the identifier of asumer and 0 = z = 10000 is the value ofcmax(u). All input numbers areegers. Except the (u,v)z triplets and the (u)z doublets, white spacan occur freely in input. Input datawhich do not terminate witain white span en

7、d of file and are correct.OutputFor each data set from the input, the progrr sumeds on the standard output thehe corresponding network. Eachum amount ofert can beresuls anegral value and is pred from the beginning of a separaine.Sle Input2 17 2132 (0,1)2013 (0,0)1(3,5)2(1,0)10 (0)15(0,1)2 (0,2)5(3,6

8、)5 (4,2)7(1)20(1,0)1(4,3)5(1,2)8 (2,3)1 (2,4)7(4,5)1 (6,0)5(0)5 (1)2 (3)2 (4)1 (5)4Sle Output156HThe s nodes,le inpontains two data sets. The er sion 0 with pmax(0)=15 anddata set encodes a network with 2 sumer 1 with cmax(1)=20, and 2er transport lines with lmax(0,1)=20 and lmax(1,0)=10. Theum valu

9、e ofis 15. The sed data set encodes the network from figure 1.SourceSoutheastern European Regional Programmingtest 2003解题思路这个题目可以采用网络流的模型来解决。在原图的基础上添加一个源点和汇点。对于每个er Sion，从源点引一条容量为pmax 的弧；对于每个sumer，引一条容量为 cmax 的弧到汇点；对于题目中给的三元组(u,v)z，从点 u 连一条容量为 z 的弧到点 v。显然这样构图是满足题目要求的。然后求这个网络的最大流即可。数据结构本题我采用的是邻接矩阵的方式来网络中的流量(fnn)和容量(cnn)，在求可改进路的时候需要用一个队列 qn和可改进路的数组 fan。其中 n 是指网络中的顶点个数，由于原图中的顶点个数不超过 100，然后添加