math 1151

......Sno_of_streets eno_of_streetsThe first line of all data set contains a positive integer no_of_intersections (greater than 0 and less or equal to 120), Which is the number of intersections in the town. The second line contains a positive integer no_of_streets, and which is the number of streets in the town. The next no_of_streets lines, one for each street in the town, is randomly ordered and represent the town ' s streets. The line corresponding to Street K (k There is no blank lines betwe

Math. Round (X)Rounding and adding 0.5 rounded downMath. Round (1.5) 2Math. Round (-11.5)-11Math. Round (-11.2)-10Math. Ceil (X)Minimum integer not less than XMath. Ceil (1.5) 2Math. Ceil (-1, 1.5)-1Math. Floor (X)Returns the largest integer less than or equal to X.Math. Floor (5.99) 5Math. Floor (-1, 5.99)-6Math. Random ()Generate random decimal places between 0 and 1Math. Random () * 7 generates random decimals between 0 and 7Math. Random () * 7 + 1

Question: Calculates the total area of the sum of N rectangles. Http://poj.org/problem? Id = 1151 Referring to the discretization pages in the black book, I thought of the polygon Filling Algorithm in computer graphics I learned last semester. SRC: /*************************************** * **************************** Created: created: 28: 3: 2013 11: 43 filename: H: \ Pe \ USA \ poj.1151.atlanta. cppfile

#1151: Domino cover issue • TwoTime Limit:10000mscase time Limit:1000msmemory LIMIT:256MBDescriptionLast week we studied the 2xN Domino problem, this week we might as well increase the difficulty, study the 3xN Domino problem?So our question is: for the 3xN chessboard, how many different coverage methods are used to cover a 1x2 Domino?First of all we can be sure that the odd length must not be covered, for even lengths, such as 2, 4, we have the follo

1151. magic board Constraints Time Limit: 1 secs, memory limit: 32 MB, special judgeDescription Questions are the same as question a. Here we expand the scope of data:N may exceed10. Please carefully consider various situations. Input The input includes multiple magic boards to be solved. Each magic board is described in three lines. N indicates the maximum number of allowed steps. The second and third lines indicate the target status. The color ran

Binary Graph matching (DFS implementation of Hungarian algorithm)Initialization: g[][] Division of vertices on both sidesBuild G[i][j] means that the i->j has a forward edge, which is the right side to the left.G is initialized to 0 without a side connectionUN is the number of vertices to match to the left, and VN is the number of vertices matching the rightCall: Res=hungary (); Output maximum number of matchesAdvantages: Suitable for dense map, DFS find augmented road, simple and easy to unders

"POJ 1151" Air RaidMinimum path coverage problem for Dag graphs (no loop directed graphs)= max Match + (n-2* Max match)(Maximum match * * For all the points that can be connectedn-2* maximum match to find all independent points plus maximum matching is the minimum required path)The following references blog: http://blog.csdn.net/whosemario/article/details/8513836 Defined: 一个有向无环图，要求用尽量少的不相交的简单路径覆盖所有的节点。 Composition: 建立一个二分图，把原图中的所

Title Address: http://ac.jobdu.com/problem.php?pid=1151Topic 1151: Bit manipulation exercisestime limit:1 seconds memory limit:32 trillion special sentence: no submission:1687 resolution:927 Title Description: Give two nonnegative integers not greater than 65535, and determine whether one of the 16-bit binary representations can be obtained by another 16-bit binary representation by looping left several bits.The differenc

HDU 1151 Air Raid (minimum path coverage)Bipartite Graph Matching (dfs implementation of the Hungary algorithm)Initialization: Division of vertices on both sides of g [] []Create g [I] [j] to indicate that the directed edge of I-> j can be used, which is the matching on the left to the right.When g is connected without edges, the initialization is 0.UN is the number of vertices on the left and vN is the number of vertices on the right.Call: res = hung

1151:0 start-up algorithm 58--open light problem time limit:1 Sec Memory limit:64 MB 64bit IO Format:%lldsubmitted:3195 accepted:1445[Submit] [Status] [Web Board] DescriptionThe computing Center has 8 computer rooms, each with n computers. Each computer has a number, such as the number of room 8th is H1 to HN, we sometimes called the H1 machine 1th, H2 for the number 2nd machine,...。One day our school ranked among the top 100 universities in

T[maxn];ll C[MAXN];structnode{ll x, y, Z; FriendBOOL operatorConstNode A,ConstNode b) {if(a.x*b.yreturn true; Else if(a.x*b.y==a.y*b.x a.zreturn true; return false; }};node ANS[MAXN];intMain () {intn=read (); intA=read (), B=read (), C=read (), d=read (); for(intI=1; i) {ans[i].z=i; ans[i].x= (ans[i-1].X*A+B)%mod+1; Ans[i].y= (ans[i-1].Y*C+D)%mod+1; } sort (ans+1, ans+1+N); LL now=0, Ans =0; for(intI=1; i) Ans+ = Now * Ans[i].y,now + =ans[i].x; printf ("%lld\n", Ans); for(intI=1; i) {

Test instructions: Given n points, your task is to make them all connected. You can create some new side, the cost equals two points distance squared (of course the smaller the better), there are several "package", you can buy, you buy, then some side can connect up,Each "package" is also spent, which allows you to find the minimum cost.Analysis: The first thought is to consider all the circumstances of the calculation, and then find the least, first Count no "package", and then some, with binar

, Num[i]) SFI (a), F[i].push_back (a); } Repu (I,1, n +1) scanf ("%d%d", xx[i], Yy[i]); Repu (i,1, n +1) Repu (J,1+ I, n +1) {W[m]= (Xx[i]-xx[j]) * (Xx[i]-xx[j]) + (Yy[i]-yy[j]) * (Yy[i]-Yy[j]); U[M]=i; V[M]=J; //printf ("%d%d%d%lf\n", M, I, J, W[m]);m++; } Repu (I,1, n +1) P[i] =i; Repu (i,0, m) R[i] =i; Sort (R, R+m, CMP); LL Minn=Kruskal1 (); Sort (KB, KB+ N-1, CMP); //printf ("%lld\n", Minn); intLim =1Q; ll cc; Repu (i,1, Lim) {CC=0; Repu (J,1, n +1) P[j] =J; Repu (J,0, q)if((1

[i][k];for (j = col+1; J A[i][col] = 0;}}return 1;}int main (){int I,j,ncas=1;int t;Double a[110][110];scanf ("%d", t);while (t--){memset (A,0.0,sizeof (a));memset (x,0.0,sizeof (x));for (int i=0;imemset (go,-1,sizeof (go));equ=var=100;int all;scanf ("%d", all);while (all--){int t1,t2;scanf ("%d%d", t1,t2);t1--;t2--;Go[t1]=t2;}for (int i=0; i{if (go[i]==-1){if (i{a[i][i]=6.0;a[i][i+1]=-1.0;a[i][i+2]=-1.0;a[i][i+3]=-1.0;a[i][i+4]=-1.0;a[i][i+5]=-1.0;a[i][i+6]=-1.0;x[i]=6;}else if (i==99){a[i][i]=

) { -Node.path.push_back (x*6+ y);//take the next step. in - if(Node.path.size () >= -){//when path has a size equal to 30, it means that it has traveled to for(inti =0; I ){ +cout 1' '; - } thecout Endl; * return true; $ }Panax Notoginseng if(Dfs (x, y))return true; -Node.path.pop_back ();//revert to original thenode.x-= _move[i][0]; +NODE.Y-=

directed edge. The side to be added when scaling. And the result should not be divided into 2 /* HDU 1151 */ # Include # Include String . H> # Include # Include Using Namespace STD; /* **************************************** * ******************************** // Bipartite Graph Matching (hungaryAlgorithm) // Initialization: G [] [] division of vertices on both sides // establish G [I] [J] to indicate the directed edge of I-> J, it is the

be matched, that is, the last point of the path. There are many points that cannot be matched, and how many paths exist. Number of paths = Original points-match the number of edges. So we make the maximum number of matching edges, that is, to minimize the number of paths. #include #includestring.h>#include#include#include#include#includeusing namespacestd;#defineINF 0X3FFFFFFF#defineMAXN 1705intN, P[maxn], Head[maxn], M, K;BOOLVIS[MAXN];structedgenode{inte, Next;} EDGE[MAXN*5];voidAddedge (intS

A very good minimum spanning tree topic. Seemingly very complex, in fact, careful analysis of the complexity of the algorithm will find that if added Lrj said optimization, in fact, the complexity is not high.As the purple book says, except for the points in the purchase package, the remaining minimum edges remain in the original minimum spanning tree. So we use the binary enumeration subset method to enumerate the combinations of all the purchase packages, then add the points in the plan to the

#include#include#includetypedefLong Longll;using namespacestd;//freopen ("d.in", "R", stdin);//freopen ("D.out", "w", stdout);#defineSspeed ios_base::sync_with_stdio (0); Cin.tie (0)#defineMAXN 2001#defineMoD 10007#defineEPS 1e-9//const int INF=0X7FFFFFFF; //infinitely LargeConst intinf=0x3f3f3f3f;/**///**************************************************************************************inline ll read () {intx=0, f=1;CharCh=GetChar (); while(ch'0'|| Ch>'9'){if(ch=='-') f=-1; ch=GetChar ();} w

......Sno_of_streets eno_of_streetsThe first line of all data set contains a positive integer no_of_intersections (greater than 0 and less or equal to 120), Which is the number of intersections in the town. The second line contains a positive integer no_of_streets, and which is the number of streets in the town. The next no_of_streets lines, one for each street in the town, is randomly ordered and represent the town ' s streets. The line corresponding to Street K (k There is no blank lines betwe