[algogeeks] Re: Delete element in a sorted array with O(1) time complexity.
impossible , ofcourse, if not implemented with a linked list. But if ur implementation permits, then you can put a suitable no. in its place to indicate deleted values, say,-1 , for an array of non negative integers. --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/algogeeks -~--~~~~--~~--~--~---
[algogeeks] Re: Why we need prime numbers?
thanks bart, and again if someone checks his/her mail frequently then its more important for him not the older mail but the newer one. So why we should not prefer top post ? --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/algogeeks -~--~~~~--~~--~--~---
[algogeeks] Re: Getting Wronge Answer
hi mukesh, thanx , i got it after all. minhaz. --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/algogeeks -~--~~~~--~~--~--~---
[algogeeks] Re: Getting Wronge Answer
hey mukesh,
this code got 3 P.E. I am not sure why. Is it helpful?
#includestdio.h
#includestring
using namespace std;
int res[20],taken[20],c_len; bool p[35];
void prime(void);
void init(void);
void dp(int);
void main(void){
int cases=0;
/* freopen(c:pr.txt,w,stdout);*/
prime();
/*/ for(c_len=1;c_len17;c_len++){*/
while(scanf(%d,c_len)==1){
// if(cases0)puts();
cases++; init();
res[0]=1;taken[1]=1;
printf(Case %d:\n,cases);
dp(1);
puts();
}
}
void dp(int n){
int i;
if(n==c_len){
if(p[res[c_len-1]+1]==1){
for(i=0;ic_len;i++){
printf(%d ,res[i]);
}
puts();
}
}
for(i=2;i=c_len;i++){
if(taken[i]==0 p[res[n-1]+i]==1){
res[n]=i;
taken[i]=1;
dp(n+1);
taken[i]=0;
}
}
}
void init(void){
memset(res[0],0,18);
memset(taken[0],0,18);
}
void prime(void){
for(int i=1;i32,p[i]=false ;i++);
p[2]=p[3]=p[5]=p[7]=p[11]=true;
p[13]=p[17]=p[19]=p[23]=p[29]=p[31]=true;
}
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[algogeeks] Re: Graph Problem
Can it be a solution? At first let us think all the edges are undirected. That is if a adjacent to b then both a-b and b-a are present. then we discard a-b if a is the current vertex with maximum outdegree and b is its adjacent with minimum outdegree and b-a is present we continue it again and again untill all the extra edges are removed. will it work? --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/algogeeks -~--~~~~--~~--~--~---
[algogeeks] Re: problem regarding lexicographic path
well mukesh ,
I solved the problem iteratively from right to left and filling up an arry.
Here is my code. I think Lago Haryanto was right. you have to select and
find out the lexicographically shortest path greedily.
Minhaz
code:
/* @JUDGE_ID: 55890 116 C++*/
#includestdio.h
void min3(int,int);
void dp(void);
void print(void);
struct datum{
long long cur_val;
long long cur_sum;
int nextro;
};
struct datum data[11][101];
int cur_ro,cur_col;
void main(){
int i,j;
while(scanf(%d,cur_ro)==1){
scanf(%d,cur_col);
for(i=1;i=cur_ro;i++){
for(j=1;j=cur_col;j++){
scanf(%lld,data[i][j].cur_val);
}
}
dp();
print();
}
}
void dp(void){
int i,j;
for(i=1;i=cur_ro;i++)
data[i][cur_col].cur_sum=data[i][cur_col].cur_val;
for(j=cur_col-1;j0;j--){
for(i=1;i=cur_ro;i++){
min3(i,j);
}
}
}
void min3(int row,int col){
int a,b,c;
long long sum_a,sum_b,sum_c;
b=row;
sum_b=data[row][col+1].cur_sum;
a=row-1;
c=row+1;
if(row==1)
a=cur_ro;
if(row==cur_ro)
c=1;
sum_a=data[a][col+1].cur_sum;
sum_c=data[c][col+1].cur_sum;
if(sum_asum_b){
if(sum_asum_c){
data[row][col].nextro=a;
data[row][col].cur_sum=sum_a+data[row][col].cur_val;
}
else if(sum_a==sum_c){
if(ac){
data[row][col].nextro=a;
data[row][col].cur_sum=sum_a+data[row][col].cur_val;
}
else{
data[row][col].nextro=c;
data[row][col].cur_sum=sum_c+data[row][col].cur_val;
}
}
else{
data[row][col].nextro=c;
data[row][col].cur_sum=sum_c+data[row][col].cur_val;
}
}
else if(sum_a==sum_b){
if(ab){
if(sum_asum_c){
data[row][col].nextro=a;
data[row][col].cur_sum=sum_a+data[row][col].cur_val;
}
else if(sum_a==sum_c){
if(ac){
data[row][col].nextro=a;
data[row][col].cur_sum=sum_a+data[row][col].cur_val;
}
else{
data[row][col].nextro=c;
data[row][col].cur_sum=sum_c+data[row][col].cur_val;
}
}
else{
data[row][col].nextro=c;
data[row][col].cur_sum=sum_c+data[row][col].cur_val;
}
}
else{
if(sum_bsum_c){
data[row][col].nextro=b;
data[row][col].cur_sum=sum_b+data[row][col].cur_val;
}
else if(sum_b==sum_c){
if(bc){
data[row][col].nextro=b;
data[row][col].cur_sum=sum_b+data[row][col].cur_val;
}
else{
data[row][col].nextro=c;
data[row][col].cur_sum=sum_c+data[row][col].cur_val;
}
}
else{
data[row][col].nextro=c;
data[row][col].cur_sum=sum_c+data[row][col].cur_val;
}
}
}
else{
if(sum_bsum_c){
data[row][col].nextro=b;
data[row][col].cur_sum=sum_b+data[row][col].cur_val;
}
else if(sum_b==sum_c){
if(bc){
data[row][col].nextro=b;
data[row][col].cur_sum=sum_b+data[row][col].cur_val;
}
else{
data[row][col].nextro=c;
data[row][col].cur_sum=sum_c+data[row][col].cur_val;
}
}
else{
data[row][col].nextro=c;
data[row][col].cur_sum=sum_c+data[row][col].cur_val;
}
}
}
void print(void){
int min_ro,i;
long long min_sum;
if(cur_ro!=0)
min_ro=1;
else
min_ro=0;
if(cur_col!=0)
min_sum=data[1][1].cur_sum;
else
min_sum=0;
for(i=2;i=cur_ro;i++){
if(min_sumdata[i][1].cur_sum){
min_sum=data[i][1].cur_sum;
min_ro=i;
}
}
printf(%d,min_ro);
for(i=1;icur_col;i++){
min_ro=data[min_ro][i].nextro;
printf( %d,min_ro);
}
printf(\n%lld\n,min_sum);
}
On 12/13/06, mukesh tiwari [EMAIL PROTECTED] wrote:
hi Lego Haryanto
i tried ur metoh but still not working ...here is my code and i m
tired with this program . i am not getting even a single method to find
lexicographic shortest path.
#includestdio.h
int min1(int k,int m,int n)
{
int min;
min=k;
if(minm)
min=m;
if(minn)
min=n;
return(min);
}
main()
{
int
m1[100][100],a[100][100],i,j,m,n,min,k,arr[100],t,min2,t1,t2,t3,v1;
while(scanf(%d%d,m,n)==2)
{
for(i=0;im;i++)
for(j=0;jn;j++)
scanf(%d,a[i][j]);
for(i=0;im;i++)
m1[i][0]=a[i][0];
for(j=1;jn;j++)
for(i=0;im;i++)
m1[i][j]=a[i][j]+min1(m1[(i+m-1)%m][j-1],m1[i][j-1],m1[(i+1)%m][j-1]);
/*for(i=0;im;i++)
{
for(j=0;jn;j++)
printf(%d,m1[i][j]);
printf(\n);
}
printf(\n);*/
min2=m1[0][n-1];
k=0;
for(i=1;im;i++)
if(m1[i][n-1]min2)
{
min2=m1[i][n-1];
k=i;
