nxt creations

space between every two integers, represents a two-way road from City u to City v, with a multiplication cost of W.The data guarantees that a tree is entered, and the root node number is 1. Output A total of one row, containing an integer, represents the minimum multiplication total cost required to control the outbreak. Input example 92 1 8881 3 882 4 65 2 86 3 1003 7 108 3 507 9 1 Output example 102

Title Link: hdu_4918_query on the subtreeTest instructionsGive a tree of n points, each point has a weight, there are two operations, one is to change the weight of a point to v, and the other is to query the distance point you do not exceed the value of the point of D and.ExercisesHere you can go to the film Bird God Blog.The simple thing is to create two tree arrays for each center of gravity of the tree, then modify each point in the bit of the center of gravity, and query the bit in each cen

The dish recommended "water problem" abused me a day t ... (Vegetables are good qwq~Obviously is a score plan problem, two points answer calculate P[i]-mid*s[i] After running on the tree depends on the backpack, choose K Maximum value if the >0 explanation also has more excellent solution.The first contact tree-dependent backpack, so the previous more than 10 FA WA and tle are all wrong, I still naive t tThe general practice of tree-dependent backpacks is to DFS sequence DP, set F[I][J] as the D

Unicomonline code are compared to God, read long time only understand QWQ. time Complexity $o (\frac{n*m}{32}$) $#include #include#include#include#include#include#defineGetChar () (p1==p2 (p2= (p1=buf) +fread (Buf,1,1Charbuf[1 at],*p1=buf,*p2=buf;using namespacestd;Const intmaxn=30001; inlineintRead () {CharC=getchar ();intx=0, f=1; while(c'0'|| C>'9'){if(c=='-') f=-1; c=GetChar ();} while(c>='0'c'9') {x=x*Ten+c-'0'; c=GetChar ();} returnx*F;}structnode{intU,V,

Long Longusing namespacestd;ConstLL maxn=2*1e7+Ten;ConstLL inf=1e7+Ten; inlineCharNC () {Static CharBuf[maxn],*p1=buf,*p2=buf; returnp1==p2 (p2= (p1=buf) +fread (buf,1, Maxn,stdin), P1==P2)? eof:*p1++;} Inline LL Read () {CharC=NC (); LL x=0, f=1; while(c'0'|| C>'9'){if(c=='-') f=-1; c=NC ();} while(c>='0'c'9') {x=x*Ten+c-'0'; c=NC ();} returnx*F;} LL MX=0; LL num[ the]={2,3,5,7, One, -, -, +}; ll Fastmul (ll a,ll b,ll p) {ll tmp= (a*b-(LL) ((Long Double) a/p*b+1e-8)*p); returntmp0? tmp+p:tmp

);}structNode {ll u,v,w,nxt;} e[200051];ll cnt,head[ -],ssw[ -],g[20005],cost[ -],times[ -];BOOLfl[ -],inch[ -],vis[ -];ll ans[20005];inlvoidAdde (ll u,ll v,ll W) {e[++cnt].u=u;e[cnt].v=v;e[cnt].w=W; E[CNT].NXT=head[u];head[u]=CNT;} ll d[ -][ the][2005],n,m;voidDP (ll x) {if(Vis[x])return; vis[x]=1; if(!fl[x]) {//Basic EquipmentTimes[x]=min (times[x],m/cost[x]); for(Re ll i=times[x];~i;i--) { f

Sample output 109 Tips in the sample, the minimum value is 109=1+2+5+7+9+12+19+21+33. Source Yuanpei-from WHF Topic links Cost Flow #include #include#defineM 100000#defineN 105#defineINF 0x3f3f3f3fusing namespacestd;BOOLInq[n];intN,s,t,cnt=1, fx[5]= {1,-1,0,0},fy[5]= {0,0,-1,1},a[n][n],to[m1],dis[m],fa[m],h

the experience of doing the problem: The character set can compress as much as possible (and discretization is one reason), trie after all | The constant of the s| is here. This optimization is very important when querying for a jump match (int v=u; v!vis[v]; v=nxt[v]) . Above.1#include 2 Const intMaxnode =10000035;3 Const intMAXN =10000035;4 5 Chars[maxn],t[103];6 intn,m,c[100035],l[100035],cnt;7 structAcautomaton8 {9 BOOLVis[

Header file/*** function: Lower push stack (push_down stack) No more memory * Time: August 18, 2014 08:13:36* Author: cutter_point*/#ifndef stack_h_included#define Stack_h_ The includedstruct stack{ struct link { void* data;//This structure is link* next;//this pointer to this structure void Initialize (void* dat, link* nxt); Initialize this structure, and the elements and next point to }*head; void Initialize ();

#include#include#defineAddedge (x, Y, z) add_edge (x, Y, z), Add_edge (y,x,0);using namespacestd;Const intmaxn=1e6+1;Const intinf=1e8+Ten; inlineCharNC () {Static CharBuf[maxn],*p1=buf,*p2=buf; returnp1==p2 (p2= (p1=buf) +fread (buf,1, Maxn,stdin), P1==P2)? eof:*p1++;} InlineintRead () {CharC=NC ();intx=0, f=1; while(c'0'|| C>'9'){if(c=='-') f=-1; c=NC ();} while(c>='0'c'9') {x=x*Ten+c-'0'; c=NC ();} returnx*F;}intn,m,s,t;structnode{intV,FLOW,NXT;}

#!/usr/bin/env python #-*-coding:utf-8-*-import requests from BS4 import beautifulsoup import bs4 import lxml def ha Ve_next (ele): Try:ele.next () Except:return False return True def is_child (Child, father): If Father:return True seek_list = father.contents for i in Seek_list:if isinstance (i , bs4.element.NavigableString): Pass elif child in I:return True else: Flag = Is_child (Child, i) if flag = = True:return True return False def Get_content_bet Ween_tables (Pre,

elements first: #include#include#include#include#include#include#include#includeUsingNamespace Std;#define INT __int64#define INF0x3f3f3f3fConstint MAXN=555;int team[MAXN];int Head[MAXN];int point[MAXN];int NXT[MAXN];int edgecnt;BOOL Vis[MAXN][maxn];int Ingrade[MAXN];voidIni(){Memset(Team,0,sizeof(Team));Memset(Head,-1,sizeof(Head));Memset(Point,-1,sizeof(Point));Memset(NXT,-1,sizeof(

edge in another graph G' with a vertex.Vertex [a]. push_back (arch (I Vertex [B]. push_back (arch (I }A = m // Note that the new edge (S, T) must be specially processed.Vertex [s]. push_back (arch (a, B, acos (-1.0 )));Vertex [t]. push_back (arch (B, a, 0 ));}Int nxt [maxm * 2], belong [maxm * 2], cnt;Void find (int x, int root ){If (nxt [x]! = Root)Find (nxt [x

Test instructions: There is a n*m of ore, each with a mine, or a portal, or a rock in the way. In addition to the rock, other squares can go right or down and walk to a non-rocky lattice. For each occurrence, it will be able to get all of its ore, and for each portal you can choose to transmit or not transmit, continue to the right or downward (portal delivery point may also be Rock), from top to bottom, from left to right in order for each portal given a delivery point. How much ore can be obta

A very common idea is to preprocess the two single-source shortest paths of s and T, and then enumerate the commercial lines. The path output can save the PA array at the shortest possible time, or it can be checked with the dist array.#include using namespacestd;Const intMAXN =502, MAXM =2002;intHEAD[MAXN], TO[MAXM], nxt[maxm],wei[maxm],ecnt;voidAddedge (intUintVintW) {to[ecnt]=v; NXT[ECNT]=Head[u]; WEI[EC

Must pay attention to the INF to open very large. Otherwise... Hey, heh.Because the traffic may cross, then b1,b2 to run again.1 A is cool.#include #include#include#include#include#defineMAXV 650#defineMaxe 250500#defineINF 9999999999999999using namespacestd;structedge{Long LongV,F,NXT;} E[maxe];Long LongG[maxv],n,a1,a2,an,b1,b2,bn,nume=1, map[ -][ -],s,t,flag;Long LongDis[maxv];queueLong Long>Q;BOOLVIS[MAXV];Charss[ -];voidAddedge (Long LongULong Lon

2 We find that useful lattices are not many, and after a detailed discussion of the classifications, only these squares are useful: Four angles, as well as obstacles (or deceleration zone) itself and the upper and lower left and right four directions, as well as the obstacles in the row (and ±1) of the tail two. Then we just have to sort all the obstacles and deceleration bands by X-Y and y-x, for each useful lattice two points to find left and right will push to where, even the end of the viol

[100010];ll f[500010];ll las[500010],ans;ll nxt[500010];p riority_queueint,int> > Q;//HeapFhakioi Main () {n=read (); K=read ();p =read (); for(LL i=1; i1;//Read in for(LL i=p;i>=1; i--) {Nxt[i]=Las[a[i]]; Las[a[i]]=i; }//the distance between pre-processing and the next occurrence for(intI=1; i){ if(use[a[i]]!=0) {Q.push (Make_pair (nxt[i],a[i));/

What's the meaning of the TMD problem? It's annoying to knock. Feel nothing to use and then knocked out ... What does it mean to be greedy for a light exam?Don't want to write anyway. Noip met the ... I don't care about him.#include #include#include#include#defineMAXN 1050#defineMAXM 10050using namespacestd;intn,m,k,t[maxm],a[maxm],b[maxm],d[maxn],mx[maxn],arr[maxn],sum[maxn],nxt[maxn],ans=0;intMain () {scanf ("%d%d%d",n,m,k); for(intI=2; i"%d",D[i])

'; c=GetChar ();} - returnret*fh; - } - + intN,m,cte,num; - intHead[n],fa[n]; + structedge{intTo,nxt;ll Val;} edge[n*2]; A at intFIND_FA (intx) { - intY=x,pre; while(Y!=fa[y]) y=Fa[y]; - while(fa[x]!=y) {pre=fa[x],fa[x]=y,x=pre;} - returny; - } - voidAeintUintVintW) { inCte++;edge[cte].to=v,edge[cte].val=W; -edge[cte].nxt=head[u],head[u]=CTE; to } + - structkrus{intx,y,flag;ll Val;} E[n]