stacked vs unstacked

implementation in Toolbox is very simple:In the NNTRAIN.M:batch_x = batch_x.* (rand (Size (batch_x)) >nn.inputzeromaskedfraction)That is, the size of the (nn.inputzeromaskedfraction) part of the X-0,denoising Autoencoder appears to be stronger than sparse autoencoderContractive auto-encoders:This variant is "Contractive auto-encoders:explicit invariance during feature extraction" proposedThis paper also summarizes a bit of autoencoder, it feels goodThe contractive autoencoders model is:whichThe

https://www.lydsy.com/JudgeOnline/problem.php?id=2809The board question WA a bit because output ans has no LLD1#include 2#include 3#include 4#include 5#include 6#include 7 using namespacestd;8 Const intmaxn=100100;9 intn,m;Ten intch[maxn][2]={},siz[maxn]={},sum[maxn]={},cnt[maxn]={},rt[maxn]={}; One intfa[maxn]={},val[maxn]={},l[maxn]={}; A inty[maxn],nex[maxn]={},head[maxn]={},tot=0; - Long Longans=0; - voidInitintXintYi) { they[++tot]=yi;nex[tot]=head[x];head[x]=tot; - } - voidUpdata (intx) {

Welcome to the original source-- Blog Park-zhouzhendong go to the blog Park to see the puzzleTopic Portal-BZOJ2809Test Instructions SummaryN points make up a tree, each point has a leadership and cost, you can make a point when the leader, and then in this point of the sub-tree select some cost and not more than M points, get leadership leadership by the number of points selected (leader can not be selected) profit. The maximum value of the profit. n≤100000 SolvingDo a tree-like DP operation.Ma

Product attributes and product specifications are the elements dynamically generated by JS. The two text boxes of product extension information are original elements, which are stacked, I thought it was because the size of the DIV where the generated element is located is not fixed, because the check boxes below the product specification are generated by the second Ajax, I doubt whether Ajax cannot push the original elements of the page down to a prop

Bootstrap Experience Instance: stacked progress bar

(); - returnnb; - } the } - intMain () {Wuyiscanf"%d", n); the for(intI=1, W; i ) { -scanf"%d", W); WuRoot[i] =NewNode (i,w); -Fat[i] =i; About } $scanf"%d", m); - for(intI=1, U,v; i ) { - Charch[Ten]; -scanf"%s", ch); A if(ch[0]=='M' ) { +scanf"%d%d", u, v); the if(Die[u] | | die[v])Continue; - intFu =find (u); $ intFV =Find (v); the if(FU==FV)Continue; theFAT[FU] =FV; theROOT[FV] =Smerge (Root[fu], ROOT[

Topic: Given n nodes, each node has an initial weight, maintaining the following operations:1. Consolidation of two unicom blocks2. A certain point weight value +x3. The right value of the Unicom block at a certain point +x4. Some weight value +x5. Ask for the weight of a point6. Ask for the maximum weight of the Unicom block at a point7. Ask for the maximum weight between all points233333333333333333333333333333333333333333333333333333333333323333333333333333333333333333333333333333333333333333

value of the node is less than or equal to the key value of its left and right child nodes.[Property 2] The distance of the left child node of the node is not less than the distance of the right child node.[Property 3] The left child node of the node is also a left-biased tree.Code#include #include#include#include#includeusing namespacestd;intRead () {intx=0, f=1;CharCh=GetChar (); while(ch'0'|| Ch>'9') {if(ch=='-') f=-1; Ch=GetChar ();} while(ch>='0' ch'9') {x=x*Ten+ch-'0'; Ch=GetChar ();} r

First look at the picture: Stacked Column chart To achieve diagonal, contrast, drag and pull effects. Large data can be reproduced in multiple dimensions. Demo Baidu Network disk download address: Http://pan.baidu.com/s/10A5Hk Process: 1, the data provided by the ASHX, serialized array of good string. The code is as follows Copy Code VarDatajs = "{ tooltip: {show:true, axispointer:

F[i] denotes the maximum value of the first I positionThen F[i]=max (F[j]) +1whichJA[j]A[i]-a[j]I-a[i]>=0J-a[j]>=0Found after two items can be launched the first item, so is an LIS problem, sorted after the tree array optimization dp can, time complexity $o (n\log N) $.#include 　　BZOJ1109: [POI2007] Stacked wood klo

Implemented function--input 1 x, add X to the small root heap, enter 2, output the minimum value and go to the heap to get rid of1 var2 I,j,k,l,m,n,head:longint;3A,lef,rig,fix:Array[0..100000] ofLongint;4 functionmin (x,y:longint): Longint;inline;5 begin6 ifX ThenMin:=xElsemin:=y;7 End;8 functionMax (x,y:longint): Longint;inline;9 beginTen ifX>y ThenMax:=xElsemax:=y; One End; A procedureSwapvarx,y:longint); inline; - varZ:

[ch[x][0]]) swap (ch[x][1],ch[x][0]);h[x]=h[ch[x][1]]+1;return x;}void dfs (int x) {sum[x]=val[x],size[x]=1;for (int i=head[x];i!=-1;i=edge[i].next) {int to=edge[i].to;Dfs (to);Sum[x]+=sum[to];size[x]+=size[to];Fa[x]=merge (Fa[x],fa[to]); }While (sum[x]>m) {sum[x]-=val[fa[x]],size[x]--;Fa[x]=merge (ch[fa[x]][0],ch[fa[x]][1]); }Ans=max (Lead[x]*size[x],ans);}int Main () {H[0]=-1;scanf ("%d%d", n,m);init ();for (int i=1;i {int f,v,l;scanf ("%d%d%d", f,v,l);if (!f) {all_root=i,val[i]

31 3 52 2 21 2 42 3 1Sample Output6HINTIf we choose a ninja numbered 1 as a manager and dispatch a third and fourth Ninja, the sum of the salaries is 4, not exceeding the total budget of 4. Since 2 ninjas were dispatched and the manager's leadership was 3,User satisfaction is 2, is the maximum number of user satisfaction can be obtained.SolutionThis is the strategy.Solution optimality: When the whole tree is DFS, we actually enumerate each manager, take the bestand solve the problem of optimali

[X] = make (l[x], y);returnx;}inline voidPopintx) {Rt[x] = make (L[rt[x]], r[rt[x]);}inline voidDfsintx) {if(!x)return; cout " ""\ n"; DFS (l[x]); DFS (R[x]);} Mainvoid) {Srand (20021214); Read (n); Read (m);intAll = n + M; for(inti =2; I for(inti = ALL; I >=2; I--) {intL =0, r =0;if(Sz[i]) { while(--sz[i]) pop (i); r = Key[rt[i]]; Pop (i); L = key[rt[i]]; Pop (i); } L = NW (l + w[i]); R = NW (r + w[i]); Rt[i] = make (Rt[i], make (L, R)); Rt[fa[i]] = make (Rt[fa[i]], rt[i]); }

({name:serisdata[i].name, Type:serisdata[i].type, Stack:serisdata[i].stack, Label: {normal: {show:true, Position:'Insideright'}}, Data:serisdat A[i].data,}); } option.series=datas; Mychart.hideloading (); mychart.setoption (option); }}, Error:function(errormsg) {alert ("Chart request data failed!"); } }); }); Script>@RequestMapping ("/showechartfoldbar") @ResponseBody Publicechar

SolSort + Tree-like array.We need to find a maximum sequence that satisfies the following conditions.\ (j-w[j]is to maintain the longest ascending subsequence of a partial-order set, and then the first and third formulas add up to the second, then the two-dimensional partial order, which is maintained by a tree array.Code/************************************************************** problem:1109 User:beiyu language:c++ Result : Accepted time:292 ms memory:2464 kb********************************

)Sample Input5100 90 66) 99 107M 1 5K 1K 1M 2 3M 3 4K 5K 4Sample Output10100066HINTSome of the data are as follows Judgeonline/upload/201607/aa.rar Naked questionLeft-leaning treeAnd check if the maintenance is in a heap.Suddenly found left to write right, but still use RC dis unexpectedly only slower than 200ms too weird#include #include#include#include#includeusing namespacestd;#defineLC T[X].L#defineRC T[X].RtypedefLong Longll;Const intn=1e6+5, inf=1e9;inlineintRead () {CharC=getchar ()

When a floating couplet advertisement is added to the page, the Flash has a stacked effect. The final cause is the onscroll event, and a setTimeout delay is added to solve the problem.Originally: Function scall (){.......}Window. onresize = scall;Window. onload = scall; Now: Function scall (){.......}Function scalltime (){Window. setTimeout ("scall ()", 1 );}Window. onscroll = scalltime;Window. onresize = scalltime;Window. onload = scalltime; SetT

]); for(intI=1; i0]=-1;scanf("%d", q); for(intI=1; iscanf('%s ', s);if(s[0]==' M ') {intx, y;scanf("%d%d", x,y);if(col[x]| | Col[y])Continue;intFx=find (x), Fy=find (y);if(FX!=FY) {intT=merge (FX,FY); fa[fx]=t,fa[fy]=t; } }Else{intXscanf("%d", x);if(Col[x])printf("%d\n",0);Else{intFx=find (x); col[fx]=1;printf("%d\n", Key[fx]); Fa[fx]=merge (ch[fx][0],ch[fx][1]); FA[FA[FX]]=FA[FX]; } } }} Copyright NOTICE: This article for Bo Master original article,

]);if(!h[ch[x][1]]) h[x]=0;Elseh[x]=h[ch[x][1]]+1;returnx;}intMain () {scanf("%d", n); for(intI=1; iscanf("%d", a[i]), a[i]-=i; for(intI=1; i1; num[tot]=1, L[tot]=r[tot]=i; while(tot>1key[root[tot]]1]]) {tot--; Root[tot]=merge (root[tot],root[tot+1]); r[tot]=r[tot+1]; num[tot]+=num[tot+1]; while(Size[root[tot]]> (num[tot]+1)/2) {Root[tot]=merge (ch[root[tot]][0],ch[root[tot]][1]); }}} ll ans=0; for(intI=1; i for(intj=l[i];jABS(Key[root[i]]-a[j]); } }printf("%lld\n", ans);} Copyright NOTICE: T