1459power network解題報(bào)告by解法或類(lèi)型網(wǎng)絡(luò)流

1、er Network解題題目來(lái)源：http解法或類(lèi)型：網(wǎng)絡(luò)流作者：/JudgeOnline/showproblem?problem_id=1459（No.1459）題目描述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解題思路這個(gè)題目可以采用網(wǎng)絡(luò)流的模型來(lái)解決。在原圖的基礎(chǔ)上添加一個(gè)源點(diǎn)和匯點(diǎn)。對(duì)于每個(gè)er Sion，從源點(diǎn)引一條容量為pmax 的弧；對(duì)于每個(gè)sumer，引一條容量為 cmax 的弧到匯點(diǎn)；對(duì)于題目中給的三元組(u,v)z，從點(diǎn) u 連一條容量為 z 的弧到點(diǎn) v。顯然這樣構(gòu)圖是滿足題目要求的。然后求這個(gè)網(wǎng)絡(luò)的最大流即可。數(shù)據(jù)結(jié)構(gòu)本題我采用的是鄰接矩陣的方式來(lái)網(wǎng)絡(luò)中的流量(fnn)和容量(cnn)，在求可改進(jìn)路的時(shí)候需要用一個(gè)隊(duì)列 qn和可改進(jìn)路的數(shù)組 fan。其中 n 是指網(wǎng)絡(luò)中的頂點(diǎn)個(gè)數(shù)，由于原圖中的頂點(diǎn)個(gè)數(shù)不超過(guò) 100，然后添加