HDU 5535 Cake structure + memory Search

Links:http://acm.hdu.edu.cn/showproblem.php?pid=5355

Test instructions: Given N and M, wherein 1<= n <= 1e5,2 <= m <= 10; Ask if there is a number of 1~n divided into M-group, so that each group is equal; if the output M line is present, each row represents a set of values, otherwise output no;

PS: A total of t<=1000 group data, or quite large;

Idea: After the pre-judgment, if possible there is a direct 2m for the period, from large to small structure and the same m-group, so that the value of n can be reduced to 2m~4m-2; because when n = 4m-1, subtract one period again, the right boundary of the next discussion is 2m-1, and it is easy to know (2m-1) *2m/2/ M is able to construct a conforming m-formula; and the pit father is, the question of the constant coefficient is not too large, the AC code runs 514MS, when the N-cycle is changed to n > 4*m-1, the direct tle; In search of the coefficient is more than one times;

DFS is also more ingenious, need to add a start to monotone find each group of data, is the final all M-group all finished and then return true, not each group completed directly return, so you can also modify the direct selection of errors ;

Determine that a group has finished simply restoring the value of the parameter to its original value, and the need to use DP optimization to set the define to shorthand;

1#include <bits/stdc++.h>2 using namespacestd;3 #defineRep0 (I,L,R) for (int i = (l); i < (R); i++)4 #defineREP1 (I,L,R) for (int i = (l); I <= (r); i++)5 #defineRep_0 (i,r,l) for (int i = (r); i > (l); i--)6 #defineRep_1 (i,r,l) for (int i = (r); I >= (l); i--)7 #defineMS0 (a) memset (A,0,sizeof (a))8 #defineMS1 (a) memset (A,-1,sizeof (a))9 #defineMSi (a) memset (A,0x3f,sizeof (a))Ten #defineINF 0x3f3f3f3f One #defineLson L, M, RT << 1 A #defineRson m+1, R, RT << 1|1 - #defineLowbit (x) (x& (×)) -typedef pair<int,int>PII; the #defineA First - #defineB Second - #defineMK Make_pair -typedefLong Longll; +typedef unsignedint UINT; -Template<typename t> + voidRead1 (T &m) A { atT x=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 ();} -m = x*F; - } -Template<typename t> in voidRead2 (T &a,t &b) {Read1 (a); Read1 (b);} -Template<typename t> to voidRead3 (T &a,t &b,t &c) {Read1 (a); Read1 (b); Read1 (c);} +Template<typename t> - void  out(T a) the { *     if(a>9) out(ATen); $Putchar (a%Ten+'0');Panax Notoginseng } - intI,j,k,n,m,l,r; the Const intN = 1e5+7; + intans[ One][n],aux[ One][n],vs[ -],ave,board; A intdp[ -][ One],rec[ -][ One][ One][ -]; the #definedef Rec[board][m] + BOOLDfsintTotintIdintstart) - { $     if(Tot = Ave) {Id++;tot =0; start =1;}//This is the end of an item, the beginning of another; of course. Returns true only when all is finished; $     if(id = =m) { -REP1 (I,1, board)if(!Vs[i]) { -aux[id][++aux[id][0]] =i; the         } -         return true;Wuyi     } the Rep1 (i,start,board) { -         if(Tot+i > Ave)return false; Wu         if(!Vs[i]) { -Vs[i] =1; Aboutaux[id][++aux[id][0]] =i; $             if(Dfs (tot+i,id,i+1))return true; -Vs[i] =0, aux[id][0]--; -         } -     } A     return false; + } the BOOLSolveinttop) - { $     if(Top >=4*m-1){ the          for(inti =1, j =0; I <= m;i++,j++){ theans[i][++ans[i][0]] = top-2*m+1+J; theans[i][++ans[i][0]] = top-J; the         } -        returnSolve (top-2*m); in     } the     Else{ theAve = top* (top+1)/m/2; About         //printf ("%d", ave); theboard =top; the         if(Dp[board][m] = =0){ the             if(Dfs (0,1,1)){ +DP[BOARD][M] =1; -REP1 (I,1, M) theRep1 (J,0, aux[i][0])BayiDEF[I][J] = Aux[i][j];//define. the             } the             ElseDP[BOARD][M] =-1; -         } -         if(Dp[board][m] = =1)return true; the         return false; the     } the } the intMain () - { the     //freopen ("Data.txt", "R", stdin); the     //freopen ("OUT.txt", "w", stdout); the     intt; read1 (t);94      for(intKase =1; Kase <= t;kase++){ the MS0 (VS); theREP1 (I,0,Ten) ans[i][0] = aux[i][0] =0; the read2 (n,m);98ll sum = 1ll*n* (n+1)/2; About         if(Sum%m | | sum/m < n | |!solve (n)) { -Puts"NO");101             Continue;102         }103Puts"YES");104REP1 (I,1, M) { theprintf"%d", ans[i][0]+def[i][0]);106Rep1 (J,1, ans[i][0])107printf"%d", Ans[i][j]);108Rep1 (J,1, def[i][0])109printf"%d", Def[i][j]); thePuts"");111         } the     }113     return 0; the}

HDU 5535 Cake structure + memory Search