图论trainningpart1F.Highways

f.highways time limit 33601000 msmemorylimit 3360漂亮的砖头64-bitintegerioformat : % lldjavaclassname 3360 maintheislandnationonofsatate y， sldmnhasaverypoorsystemofpublichighways.thesldmnngovernmentisawareofthisproblemandhasalreadyconstructedanumberofhighwaysconconononays rtanttowns.However，therearestillsometownsthatyoucan ' treachviaahighway.itisnecessarytobuildmorehighwayssothatitwilbepoucan

sldmntownsarenumberedfrom1tonandtownihasapositiongivenbythecartesiancoordinates (Xi， yi ).eachighwayconnectsexacltytwotowns.all highways (boththeoriginalonesandtheonesthataretobebuilt ) folllow straight lines ays esiandistancebetweentowns.allhighwayscanbeusedinbothdirections.highwayscanfreelycrosseachotth butadrivercanonlyswitchbetwetwethbetwetwethbetwetwethetwetwetetwethbetwetwetwetet

thesldmnngovernmentwantstominimizethecostofbuildingnewhighways.however， theywanttoguaranteethateverytownishighway-reachablefromeveryothertown.sincesldmnissoflat，thecostofahighwayisalwaysproporat theleastexpensivehighwaysystemwillbetheonethatminimizesthetotalhighwal

inputtheinputconsistsoftwoparts.thefirstpartdescribesalltownsinthecountry，andthesecondpartdescribesallofthehighwaysthathavave

thefirstlineoftheinputfilecontainsasingleintegern (1=n=750 )，representingthenumberoftowns.thenextnlineseachcontaintwow xiandyiseparatedbyaspace.thesevaluesgivethecoordinatesofithtown (for ifrom1ton ).coordinateswillhaveanabsolutevaluenogreatesogreated

thenextlinecontainsasingleintegerm (0=m=1000 )， representingthenumberofexistinghighways.thenextmlineseachcontainapairofintegersseparatedbyaspace.thesetwointegersgiveapapair onnectedbyahighway.eachpairoftownsisconnectedbyatmostonehighway。

outputwritetotheoutputasinglelineforeachnewhighwaythatshouldbebuiltinordertoconnectalltownswithminimalposssibletotalllllengthof new ldbepresentedbyprintingtownnumbersthatthishighwayconnects，separated by a space。

ifnonewhighwaysneedtobebuilt (alltownsarealreadyconnected )，theoutputfileshouldbecreatedbutitshouldbeeempty。

sample input 91500324551045212533139712 sample output 163749578

解决问题：找到最小生成树的父节点。 。

1 # include iostream2# include cstdio3# include cstring4# include cmath5# include algorithm6# include climits7# include vector 8 include string 11 # includeset 12 # definelllonglong 13 # define INF0x3F3 f14 usingnamespacestd 15 const int maxn=800； 16 int mp[maxn腼腆的故事maxn] 17 struct pos {18 int x，y； 19 ) p[maxn]； 20 int n，m，d[maxn]，pre[maxn]； 1intdis(constposa，const pos b ) 22返回(a.x-b.x ) ) a.x-b.x ) ) a.y-b.y ) ) a.y-b.y )； 23 ) 24voidprim(intsrc ) {25 int i，j，index，theMin； 26for(I=1； i=n； I({27pre[I]=src； 28 d[i]=mp[src腼腆故事i] 29 ) 30pre[src]=-1； 31for(I=1； i n； I ) {32 theMin=INF； 33索引=-1； 34for(j=1； j=n； j ) 35if(pre[j]！=-1d [ j ] them in (them in=d [ index=j ]； 36 ) 37if(themin！=0) printf('%d%d(n )，index，pre[index] )； 38 pre[index]=-1； 39for(j=1； j=n； j () 40if(pre[j]！=-1d [ j ] MP [索引腼腆故事j ] ({ 41d [ j ]=MP [索引腼腆故事j ] 42 pre [ j ]=索引； 43(45 ) 46 ) 47intmain ) ) {48 int i，j，u，v； 49Scanf('%d '，n )； 50for(I=1； i=n； I )扫描(' % d % d )、p[i].x、p[i].y )； 51for(I=1； i=n； I ) 52for(j=I1； j=n； j ) 53 mp[i腼腆故事j]=mp[j腼腆故事I]=dis(p[I]，p[j]； 54 ) 55扫描(' % d '，m )； 56while(m-- ) 57scanf ) ' %d%d '，u，v )； 58 mp[u腼腆故事v]=mp[v腼腆故事u]=0； 59 ) 60prim(1； 61返回0； 62 )视图代码