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HUST 1017 Exact cover (DLX), hustdlx DescriptionThere is an N * M matrix with only 0 s and 1 s, (1 InputThere are multiply test cases. first line: two integers N, M; The following N lines: Every line first comes an integer C (1 OutputFirst output the number of rows in the selection, then output the index of the selected rows. if there are multiply selections, you shoshould just output any of them. if there are no selection, just output "NO ". Sample

Analysis: Open an array of 300w, statistics, and then NLOGN statistics for each value in order to the first occurrence of the number of casesTime complexity: O (NLOGN) n at 3e6 order of magnitude#include #include#include#include#include#includeusing namespaceStd;typedefLong LongLL;Const intN =3e6;ConstLL mod = 1e9+7; LL N,m,sum[n+5],c[n+5];intMain () {scanf ("%i64d",N); intmx=0; for(intI=1; ii) { intX;SCANF ("%d", x); + +C[x]; MX=Max (X,MX); } for(intI=1; ii) { for(intj=1; jj) {

First to analyze: First time: 150Second time: 150/4 = 37 2Third time: (37+2)/4 = 9 3Fourth time: (9 + 3)/4 = 3 3 Initial value 150 bottle of water, which is 150 caps1. Take all the lids to change the water, record the remainder and sum2. (last remainder + last change of water to drink)/4, record remainder and sum1......Infinite loops, until the number of lids Binding code: Public classDemo3 { Public Static voidMain (string[] args) {intn = 150; intsum = 150; while(N >= 4) {　　　

*A.A)/(a.a*b.b-a.b*B.A));} Lne zcx (PNT a,pnt b) {db MX= (a.x+b.x)/2, my= (A.Y+B.Y)/2; DB La=a.x-b.x,lb=a.y-b.y,lc=-(mx*la+my*lb); returnLne (LA,LB,LC);} PNT wx (PNT a,pnt b,pnt c) {returnZcx (A, b) *zcx (a,c);} DB EPS=1e-6;p NT NP (PNT A,PNT b,pnt c) {if(Fabs (b-a) * (c-a)) EPS) { DoubleAbl=dis (A, B), Acl=dis (A,c), Bcl=dis (b,c), ans=-1; PNT ap; if(Abl>ans) Ans=abl, ap=Mid (A, b); if(Acl>ans) Ans=acl, ap=mid (A,c); if(Bcl>ans) ANS=BCL, ap=mid (B,c); returnap; } returnWX (a,b,

= n+m+2; the for(inti = s; I ) g[i].clear (); the DoubleVal;94 for(inti =1; I ) the { thescanf"%LF", val); the Addedge (S, I, log (val));98 } About for(inti =1; I ) - {101scanf"%LF", val);102Addedge (i+M, T, log (val));103 }104 for(inti =1; I ) the {106 intA, B; scanf"%d%d", a, b);107Addedge (A, B +M, INF);108 }109 DoubleFlow =Maxflow (); theprintf"%.4lf\n", exp (flow));111 } the

if(p==-1|| Sum[p]>Sum[i]) -p=i; in Delete (p); the for(intI=d[p];i!=p;i=D[i]) { theans[dep]=Row[i]; About for(intj=r[i];j!=i;j=R[j]) Delete (Col[j]); the if(Solve (dep+1)) the return true; the for(intj=l[i];j!=i;j=L[j]) Resume (Col[j]); + } - Resume (p); the return false;Bayi } the the }DLX; - intMain () - { the while(SCANF ("%d%d", n,m) = =2) the { the DLX. Init (n,m); the for(intI

P1057 Build HouseLabel not submitted: [Show label] DescriptionThe eternal Soul has recently been given an area of n*m (happy ing ^_^), he wants to build a house on this land, the house must be square.However, this land is not perfect, there are many uneven places (can also be called flaws). These flaws are so disgusting that they cannot be brick by brick on top.He wanted to find the biggest square of the land to build the house with no blemish.However, this is not a problem, the eternal soul in

} About } the the BOOLDancingintk) the { + intc =R[head]; - if(c = =HEAD) the {Bayi output (); the return true; the } - - Remove (c[c]); the for(inti = d[c];i! = C;i =D[i]) the { theANS[K] =H[i]; the for(intj = r[i];j! = I;j =R[j]) - Remove (c[j]); the if(Dancing (k +1)) the return true; the for(intj = l[i];j! = I;j =L[j])94 Resume (c[j]); the } the Resume (c[c]); the 98 return false; About } - 101 voi

operator* (ConstPoint a) {returnX*A.X+Y*A.Y;}Double operator^ (ConstPoint a) {returnx*a.y-y*a.x;}} Pt[n];structline{Point P,v; Line () {A,point b):p (a), V (b) {}};structcircle{point P;DoubleR Circle () {} circle (Point _p,Double_r):p (_p), R (_r) {}}ans;DoubleDis (point a,point b) {return sqrt(A-B) * (a));BOOLIs_in_cir (Point p,circle c) {returnDis (P,C.P) Doubletmp= (l2.v^u)/(L1.V^L2.V);returnl1.p+l1.v*tmp; }point rotate (Point a) {returnPoint (-a.y,a.x);}intNintMain () {Srand (19990329);scan

, expand (The detailed one) function and stop the event stream (that is, click on the vertical, expand the detailed one)document.getElementById (' Status_show '). AddEventListener (' Click ', Stopevent, false);Because false determines the event bubbling, in order to prevent point details, bubbles up to document, triggering the click of the document setting Hidebox method, in Status_show we want to bind the block event stream function.}This understand, for IE and then write a attachevent can, of

Ideas see Hihocoder, with the Kuangbin matrix fast Power, a ac,6 of a stroke.#include 　　Hihocoder 1151 Domino cover problem two (Matrix fast power)

The logic cover method of white box test A white box test is also called a logical drive test or a structural test. Differences from black Box testing: focus on the internal structure of the code The premise of White box testing: The internal logic of the known code Test method: Logic Overlay method The logic covering method consists of six kinds, from weak to strong: statement overlay, decision coverage,

Test instructionsGive you n a bag, want to get M something (number 1~m each can have only one)Then there are k items in each bag, and each item has a number.Ideas:Dance even template problemCode:#include "stdio.h" #include "algorithm" #include "string.h" #include "iostream" #include "queue" #include "map" #include " Vector "#include" string "using namespace std; #define N 1010*100//Total element number # # RN 1010//Total number of rows # define CN 1010// Total number

The minimum point coverage problem of general graphs is a NPC problem, and there is no better polynomial algorithm for ha. But this problem has a little special place, each side must contain the first 30 points of one, so this question can enumerate the money 30 points of the selected and not selected state, the subsequent point corresponding to the state is unique. So this is the dfs+ feasibility of the reduction of the branch and the optimal reduction of the answer.#include SCU OJ 4439:vertex

(Constpoint C,intID) - { A for(inti = ID +1; I i) + { the DoubleD = Length (C-p[i]); - if(DCMP (d-radius[i]) 0)return false;//the key to this problem $ } the return true; the } the the intMainvoid) - { in //freopen ("10969in.txt", "R", stdin); the intT; thescanf"%d", T); About while(t--) the { thescanf"%d", n); the for(inti =0; I "%LF%LF%LF", radius[i], p[i].x, p[i].y); + - Doublesum =0.0; the for(inti =0; I i

Http://www.cnblogs.com/grenet/p/3145800.htmlBest elaboration on Dancing list I found. In Chinese.Traditional backtracing is a heuristic-like procedure. Say, there was N items to try with one by one, when N[i] was picked, all rest n[i + 1, N-1] items would being tested one by E, too. What's more, due to recursive calls, there is a lot intermediate caching necessary to contain intermediate states. Very blindly procedure correct?Traditional backtracing is like, you are trying to find a place on the

Why does my panel always cover SpeedButton? Is there a problem if transparent of my SpeedButton is set to true? But it cannot be changed to false,I am a cainiao. I posted my post for the first time. Help me! Hedong(Hedong) ▲▲▲▲△ - Box active member 2003-4-28 13:24:43 8 th floor: is this a problem? You need to read the ent

The title is a very obvious two-dimensional graph with weighted matching model,Add source point to NX build edge, NY to meeting point build edge (NX. NY) =inf The edge of the building. The minimum cut is the minimum point weight coverage.In the subject because of the product, so all the log is converted to addition, and finally back.#include poj3308 Paratroopers---Minimum point weight cover-gt; min cut

Use JQ to cover the single-screen page rolling effect of Baidu books pre-sale pageThe principle page rolling method must be implemented through animate. The clicked a tag index is used to determine which page to roll. At the same time, we also need a global variable to store the index of the current page, so that we can roll the page through the scroll wheel event. In addition, the callback event is also called here, including the callback of the acti

Pop up a div to set the background color and transparency of the DIVRemove this div or hide it when loading is completeAs for fading through settimeout or setinterval, try to change the transparency..test{width:100%;height:100%;position:absolute;background-color:#5c5c5c;z-index:100;display:none;filter:Alpha(Opacity=40,finishOpacity=40,style=0,startX=100,startY=100,finishX=100,finishY=100);}这个是你说的灰色透明的背景，透明度什么的可以在那个滤镜中修改。至于你说的淡入淡出效果就用setTimeout来做。这个方法应该是显示这个界面和登陆成功后隐藏这个界面时条用。JS implementation loa