allscripts adp

= g[u].size ()-1; I >=0; -I.) SON[U] + =DFS (G[u][i]); - sort (G[u].begin (), G[u].end (), CMP); + for(inti =0; I i) { Adp[u][1] + = dp[u][0]*son[g[u][i]] + dp[g[u][i]][1] +Son[g[u][i]]; atdp[u][0] + = dp[g[u][i]][0] +2; - } - if(Worms[u]) dp[u][0] =0; - returnSon[u]; - } - intMain () { in Charstr[5]; - while(SCANF ("%d",N), N) { to for(inti =1; I i) g[i].clear (); +Memset (DP,0,sizeofDP); -Memset (Worms,false,sizeo

Use dp[i][j] to indicate the first I task, a complete need J time, B completed need dp[i][j] time.The recursive equation is also easy to get outDp[i][j]=min (Dp[i-1][j]+b[i],dp[i-1][j-a[i])//Where A[i],b[i] respectively indicates the time required for A/b to complete the task I.Attached code1#include 2#include 3#include 4#include 5#include 6#include 7 using namespacestd;8Ifstream Fin ("divide.in");9Ofstream Fout ("Divide.out");Ten intdp[203][40005]={0},rws=0, rw[203][2]={0}; One intMainvoid) A {

QuestionGiven a 2D binary matrix filled with 0 's and 1 ' s, find the largest square containing all 1 's and return its area.For example, given the following matrix:1 1 1 1 11 0 0 1 0Return 4.Solution2-d Array. The changing condition is:T[i][j] = min (t[i][j-1], t[i-1][j], t[i-1][j-1]) + 1.1 Public classSolution {2 Public intMaximalsquare (Char[] matrix) {3 if(Matrix = =NULL|| Matrix.length )4 return0;5 introw = matrix.length, column = matrix[0].length, max = 0;

Chairman Tree1#include 2#include 3#include 4 #defineN 20000105 using namespacestd;6mapint,int>Ma;7 intn,m,i,a,b,c,e[n],last,d[n],id[n],deep[n],j;8 intDp,pre[n],p[n],tt[n],l[n],r[n],ls[n],rs[n],s[n],tot,root[n];9 intjump[100100][ +];Ten voidLinkintXinty) One { Adp++;p re[dp]=p[x];p [x]=dp;tt[dp]=y; - } - voidBuildintx,intAintb) the { -tot++;x=tot; -l[x]=a;r[x]=b; - if(b-a>1) + { - intM= (a+b) >>1; + build (ls[x],a,m); A build (rs[x],m,b

to come to the point of meeting and continue to the right, the lower left corner of the person can move up to meet the point of continued upstream or The upper-left corner of the person moves down to the meeting point and continues to move down, the lower left corner of the person to the right to meet the point to continue to move to the right as the maximum value simple DP1#include 2 using namespacestd;3 Const intMAXN =1005;4 intdp[4][maxn][maxn],a[maxn][maxn],n,m;5 intMain () {6 while(~S

the tree with this node as the root node is less than or equal to n/2. The node is a node that can be deleted;1#include 2#include 3#include 4#include 5#include 6#include 7 using namespacestd;8 Const intmaxn=10007;9 intDP[MAXN];Ten BOOLINPUT[MAXN]; Onevectorint>G[MAXN]; A intN; - voidInit () - { theMemset (DP,0,sizeof(DP)); -memset (Input,0,sizeof(input)); - for(intI=0; i) - g[i].clear (); + } - voidBFsintPointintfather) + { Adp[point]=1; at

these two subsequent states is the number of scenarios for the previous state.Of course, the next is to give the two-dimensional DP to the initial value, obviously dp[1][0]=1, and then found in fact dp[1][i]=i+1 (can be understood: when the jar is only a half-shaped pill and 1 tablets of complete pills, for the number of solutions, in fact, the main thing is to see the whole pill in the first few days to eat, For a total of i+1 days, it is clear that the number of C (i+1) (1) programs. Then I i

Given a 2D binary matrix filled with 0 's and 1 ' s, find the largest square containing all 1 's and return its area.For example, given the following matrix:1 1 1 1 11 0 0 1 0In the given two-dimensional character array, find out the number of 1 that the rectangle with the maximum total of 1 contains:Note that this is the equivalent of the area, only ask for the side length is very good to run. Side length with DP very good, this problem in fact with the former Youdao DP problem very much like,

| | j==0) inju[i][j]=1; - ElseJu[i][j]= (ju[i-1][j]+ju[i-1][j-1])%N; to } + } -scanf"%d",k); the ints; * intn,m; $ intx, y;Panax Notoginseng for(s=1; s) - { the for(i=0; i1050; i++) + { Adp[i]=1; the vec[i].clear (); + } -scanf"%d",n); $ for(i=1; i) $cnt[i]=i; - for(i=1; i) -

[ $][ -]; - intMain () { the intN; +scanf"%d",n); Amemset (dp,-1,sizeof(DP)); the intTotal =0 ; + for(inti =1; I ) - { $scanf"%d%d",a[i].t,A[I].W); $Total + =a[i].t; - } -dp[0][0] =0 ; the intMi =1e9; - for(inti =1; I )Wuyi { the for(intj =0; J ) - { WuDP[I][J] = dp[i-1][j]; - } About for(intj =0; J ) $ { - if(dp[i-1][J]! =-1) - { - if(dp[i-1][J+A[I].T] = =-1)

DescriptionFarmer John commanded his cows to search forDifferent sets of numbers that sum to a given number. The cows use is numbers that is an integer power of2. Here is the possible sets of numbers that sum to7: 1)1+1+1+1+1+1+1 2)1+1+1+1+1+2 3)1+1+1+2+2 4)1+1+1+4 5)1+2+2+2 6)1+2+4Help FJ count all possible representations forA given integer N (11, the, the).InputA single line with a single integer, N.Output as This 9 digits (inbase representation).Sample Input7Sample Output6SourceUsaco 2005 J

(I,1, N) a[i]=getint (); toF (I,1, N) { +b[i]=A[i]; -w[i][i]=0; theF (j,i+1, N) { *w[i][j]=w[i][j-1]+a[j]*B[i]; $b[i]+=A[j];Panax Notoginseng } - } theF (I,1, N) F (J,1, m) dp[j][i]=INF; +F (I,1, N) { Adp[1][i]=w[1][i]; thes[1][i]=0; + } -F (I,2, M) { $s[i][n+1]=N; $ D (j,n,i) -F (k,s[i-1][j],s[i][j+1]) - if(dp[i-1][k]+w[k+1][j]Dp[i][j]) { thes[i][j]=K; -dp[i][j]=dp[i-1][k]+w[k+1][j];Wuyi

of "49" in the 1~n closed intervalExercisesDp[i][2] Length I contains the number of "49"Dp[i][1] length I does not contain "49" but the number of "9" highDp[i][0] The length of I does not contain the number of "49"Array A[i] stores every digit of n from low to high.DP[I][2]=DP[I-1][2]*10+DP[I-1][1]; Consider the "4" i-1 bit "9" for the I-bitDP[I][1]=DP[I-1][0];DP[I][0]=DP[I-1][0]*10-DP[I-1][1];Why do you want to increase your self by 1 before n processingBecause the problem is that the closed i

using namespacestd;6typedefLong Longll;7 Const intmaxn= One;8ll dp[2][1MAXN];9 Ten voidSolveintMintN) One { A if(mN) swap (m,n); - intCur=0; -Memset (DP,0,sizeof(DP)); thedp[cur][(11]=1; - for(intI=1; i) - for(intj=1; j) - { +cur^=1; - /*cur to be placed behind the second cycle, written in the triple loop.*/ +memset (Dp[cur],0,sizeof(Dp[cur])); A /*don't forget to clear the current state*/ at for(intk=0; K11; k++) -