I participated in a programming competition at my University. I solved all the questions except this one. Now I am practicing this question to improve my skills. But I can't figure out the algorithm. If there is any algorithm existing please update me. Or any similar algorithm is present then please tell me I will change it according to this question.
This is what I want to do.
The First line of input is the distance between two points.
After that, each subsequent line contains a pair of numbers indicating the length of cable and quantity of that cable. These cables are used to join the two points.
Input is terminated by 0 0
Output:
The output should contain a single integer representing the minimum number of joints possible to build the requested length of cableway. If no solution possible than print "No solution".
Sample Input
444
16 2
3 2
2 2
30 3
50 10
45 12
8 12
0 0
Sample Output
10
Thanks guys. I found a solution from "Perfect subset Sum" problem and then made a few changes in it. Here's the code.
#include <bits/stdc++.h>
using namespace std;
bool dp[100][100];
int sizeOfJoints = -1;
void display(const vector<int>& v)
{
if (sizeOfJoints == -1)
{
sizeOfJoints = v.size() - 1;
}
else if (v.size()< sizeOfJoints)
{
sizeOfJoints = v.size() - 1;
}
}
// A recursive function to print all subsets with the
// help of dp[][]. Vector p[] stores current subset.
void printSubsetsRec(int arr[], int i, int sum, vector<int>& p)
{
// If sum becomes 0
if (sum == 0)
{
display(p);
return;
}
if(i<=0 || sum<0)
return;
// If given sum can be achieved after ignoring
// current element.
if (dp[i-1][sum])
{
// Create a new vector to store path
//vector<int> b = p;
printSubsetsRec(arr, i-1, sum, p);
}
// If given sum can be achieved after considering
// current element.
if (sum >= arr[i-1] && dp[i-1][sum-arr[i-1]])
{
p.push_back(arr[i-1]);
printSubsetsRec(arr, i-1, sum-arr[i-1], p);
p.pop_back();
}
}
// all subsets of arr[0..n-1] with sum 0.
void printAllSubsets(int arr[], int n, int sum)
{
if (n == 0 || sum < 0)
return;
// If sum is 0, then answer is true
for (int i = 0; i <= n; i++)
dp[i][0] = true;
// If sum is not 0 and set is empty, then answer is false
for (int i = 1; i <= sum; i++)
dp[0][i] = false;
// Fill the subset table in botton up manner
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= sum; j++)
{
if(j<arr[i-1])
dp[i][j] = dp[i-1][j];
if (j >= arr[i-1])
dp[i][j] = dp[i-1][j] ||
dp[i - 1][j-arr[i-1]];
}
}
if (dp[n][sum] == false)
{
return;
}
// Now recursively traverse dp[][] to find all
// paths from dp[n-1][sum]
vector<int> p;
printSubsetsRec(arr, n, sum, p);
}
// Driver code
int main()
{
int input[2000];
int inputIndex = 0;
int i = 0;
int distance = 0;
cout<< "Enter Input: " <<endl;
cin>> distance;
while(true)
{
int temp1 = 0;
int temp2 = 0;
cin>> temp1;
cin>> temp2;
if (temp1 == 0 && temp2 == 0)
{
break;
}
for (i = 0; i < temp2; i++)
input[inputIndex++] = temp1;
}
cout<< "Processing output. Please wait: " <<endl;
printAllSubsets(input, inputIndex, distance);
if(sizeOfJoints != -1)
cout<<sizeOfJoints;
else
cout<<"No Solution Possible";
return 0;
}
Good morning, Stack Overflow. You guys helped me out on an earlier assignment, and I'm hoping to get a little help on this one.
It's a programming assignment relating to sorts, one part of which is to write a working implementation of merge sort.
I adapted my solution from the pseudocode the professor used in class, but I'm getting an annoying segfault at the indicated location.
This method is sorting an array of structs, with data_t defined as struct pointers.
The struct definition:
typedef struct {
int id;
int salary;
} employee_t;
typedef employee_t* data_t;
They're being sorted by salary, which is a randomly generated number from 40,000 to 90,000.
Here's the actual method
void merge_sort(data_t items[], size_t n)
{
if (n < 2)
return;
size_t mid = (n / 2);
data_t *left = malloc(sizeof(data_t) * mid);
data_t *right = malloc(sizeof(data_t) * (n - mid));
for (int y = 0; y < mid; y++)
{
left[y] = items[y];
}
for (int z = mid; z < n; z++)
{
right[z] = items[z];
}
merge_sort(left, mid);
merge_sort(right, (n - mid));
size_t l, r, i;
l = 0;
r = 0;
for (i = 0; i < (n - 1); i++)
{
if ((l < mid) && ((r >= (n - mid)) || ((left[l]->salary) <= (right[r]->salary))))
{
items[i] = left[l++];
}
else
{
items[i] = right[r++];
}
}
free(left);
free(right);
}
Note that I haven't made it as far as the end, so the array frees might be incorrectly located.
The segfault always occurs when I try to access right[r]->salary, so I'm assuming this is related to a null pointer, or similar. However, I'm extremely new to sorting, and I don't know exactly where to properly implement a check.
Any advice is appreciated greatly.
At first glance there's this fix:
for (int z = mid; z < n; z++)
{
right[z-mid] = items[z];
}
This is a piece of Wikipedia's sample implementation of interpolation search:
public int interpolationSearch(int[] sortedArray, int toFind){
// Returns index of toFind in sortedArray, or -1 if not found
int low = 0;
int high = sortedArray.length - 1;
int mid;
while (sortedArray[low] <= toFind && sortedArray[high] >= toFind) {
mid = low +
((toFind - sortedArray[low]) * (high - low)) /
(sortedArray[high] - sortedArray[low]); //out of range is possible here
if (sortedArray[mid] < toFind)
low = mid + 1;
else if (sortedArray[mid] > toFind)
// Repetition of the comparison code is forced by syntax limitations.
high = mid - 1;
else
return mid;
}
if (sortedArray[low] == toFind)
return low;
else
return -1; // Not found
}
Why is out of range possible there?
EDIT: added the whole code
Out of range is possible here if an integer overflow occurs.
To overcome this you can you float/double rather then a 32 bit integer.
Have a look at this,
What's wrong with this Interpolation search implementation?
There is an interesting game named one person game. It is played on a m*n grid. There is an non-negative integer in each grid cell. You start with a score of 0. You cannot enter a cell with an integer 0 in it. You can start and end the game at any cell you want (of course the number in the cell cannot be 0). At each step you can go up, down, left and right to the adjacent grid cell. The score you can get at last is the sum of the numbers on your path. But you can enter each cell at most once.
The aim of the game is to get your score as high as possible.
Input:
The first line of input is an integer T the number of test cases. The first line of each test case is a single line containing 2 integers m and n which is the number of rows and columns in the grid. Each of next the m lines contains n space-separated integers D indicating the number in the corresponding cell
Output:
For each test case output an integer in a single line which is maximum score you can get at last.
Constraints:
T is less than 7.
D is less than 60001.
m and n are less than 8.
Sample Input:
4
1 1
5911
1 2
10832 0
1 1
0
4 1
0
8955
0
11493
Sample Output:
5911
10832
0
11493
I tried it but my approach is working very slow for a 7x7 grid.I am trying to access every possible path of the grid recursively and comparing the sum of every path.Below is my code
#include<iostream>
#include <algorithm>
#include <stdio.h>
using namespace std;
int max(int a,int b,int c, int d)
{
int max = a;
if(b>max)
max = b;
if(c>max)
max = c;
if(d>max)
max = d;
return max;
}
int Visit_Component( int (*A)[8], int Visit[8][8], int m,int n , int row, int col)
{
if ( ( row >= m ) || (col >= n ) || (col < 0) || (row < 0) || A[row][col] == 0 || Visit[row][col] == 1 )
{
return 0;
}
else
{
Visit[row][col] = 1;
int a= 0,b=0,c=0,d=0,result =0;
a = Visit_Component( A, Visit,m,n, row+1, col);
b = Visit_Component( A, Visit,m,n, row, col +1);
c = Visit_Component( A, Visit,m,n, row, col -1);
d = Visit_Component( A, Visit,m,n, row-1, col );
Visit[row][col] = 0;
result = A[row][col] + max(a,b,c,d);
return result;
}
}
int main(){
int T;
scanf("%d",&T);
for(int k =0; k<T;k++)
{
int N ;
int M;
int count = 0;
int maxcount = 0;
scanf("%d %d",&M,&N);
int C[8][8];
int visit[8][8];
for(int i = 0; i < M; i++)
for(int j = 0; j < N; j++)
{
scanf("%d",&C[i][j]);
visit[i][j] = 0;
}
for( int i= 0 ; i< M ; i++ )
{
for( int j =0; j< N ; j++ )
{
count = Visit_Component( C, visit,M,N, i, j);
if(count > maxcount)
{
maxcount = count;
}
}
}
printf("%d\n",maxcount);
}
return 0;
}
Please suggest me how to optimize this approach or a better algorithm.
As Wikipedia article on Travelling salesman problem suggests, there are exact algorithms, solving this task quickly. But it is hard to find any. And they are, most likely, complicated.
As for optimizing OP's approach, there are several possibilities.
It's easier to start with simple micro-optimization: condition Visit[row][col] == 1 is satisfied with highest probability, so it should come first.
Also it is reasonable to optimize branch-and-bound algorithm with dynamic programming to avoid some repeated calculations. Memorizing calculation results in simple hash table for the cases of up to 19 visited cells improves performance by more than 25% (and more may be expected for some improved hash table). Here is the modified code snippet:
#include<iostream>
#include <algorithm>
#include <stdio.h>
using namespace std;
int max(int a,int b,int c, int d)
{
int max = a;
if(b>max)
max = b;
if(c>max)
max = c;
if(d>max)
max = d;
return max;
}
typedef unsigned long long ull;
static const int HS = 10000019;
static const int HL = 20;
struct HT {
ull v;
int r;
int c;
};
HT ht[HS] = {0};
int Visit_Component(
int (*A)[8], ull& Visit, int m,int n , int row, int col, int x)
{
if ( (Visit & (1ull << (8*row+col))) || ( row >= m ) || (col >= n ) ||
(col < 0) || (row < 0) || A[row][col] == 0)
{
return 0;
}
else
{
if (x < HL)
{
HT& h = ht[(Visit+4*row+col)%HS];
if (h.v == Visit && h.r == row && h.c == col)
return 0;
}
Visit |= (1ull << (8*row+col));
int a= 0,b=0,c=0,d=0,result =0;
a = Visit_Component( A, Visit,m,n, row+1, col, x+1);
b = Visit_Component( A, Visit,m,n, row, col +1, x+1);
c = Visit_Component( A, Visit,m,n, row, col -1, x+1);
d = Visit_Component( A, Visit,m,n, row-1, col , x+1);
Visit &= ~(1ull << (8*row+col));
result = A[row][col] + max(a,b,c,d);
if (x < HL)
{
HT& h = ht[(Visit+4*row+col)%HS];
h.v = Visit;
h.r = row;
h.c = col;
}
return result;
}
}
int main(){
int T;
scanf("%d",&T);
for(int k =0; k<T;k++)
{
int N ;
int M;
int count = 0;
int maxcount = 0;
scanf("%d %d",&M,&N);
int C[8][8];
ull visit = 0;
for(int i = 0; i < M; i++)
for(int j = 0; j < N; j++)
{
scanf("%d",&C[i][j]);
}
for( int i= 0 ; i< M ; i++ )
{
for( int j =0; j< N ; j++ )
{
count = Visit_Component( C, visit,M,N, i, j, 0);
if(count > maxcount)
{
maxcount = count;
}
}
}
printf("%d\n",maxcount);
}
return 0;
}
And much more improvements may be done by pre-processing the input matrix. If there are no zeros in the matrix or if there is only one zero in the corner, you may just sum all the values.
If there is only one zero value (not in the corner), at most one non-zero value should be excluded from the sum. If you invent an algorithm, that determines the subset of cells, from which one of the cells must be removed, you can just select the smallest value from this subset.
If there are two or more zero values, use branch-and-bound algorithm: in this case it is about 20 times faster, because each zero value in input matrix means approximately fivefold speed increase.
One optimization that I can think of is to apply Dijkstra's algorithm. This algorithm will give you a minimum (in your case maximum) path for a particular source node to all destination nodes.
In this example, the first step would be to build a graph.
And because you don't know the source node to start at, you will have to apply Dijkstra's algorithm for each node in the grid. The time complexity will be better than your recursion method because for a particular source node, when finding a maximum path Dijkstra's algorithm does not go through all the possible paths.
#include<iostream>
#include<vector>
using namespace std;
vector<vector<int> >A;
vector<vector<bool> >test;
vector<vector<bool> >test1;
int sum_max=0;
int m,n;
vector<vector<bool> > stamp;
void color1(int i,int j,vector<vector<bool> >temp_vector,vector<vector<bool> > st,int summ){
temp_vector[i][j]=false;summ+=A[i][j];st[i][j]=true;
//1.1
if(i+1<m && temp_vector[i+1][j]){
if(test1[i+1][j]){
if(sum_max<(summ)){sum_max=summ;stamp=st;}
}
else{color1(i+1,j,temp_vector,st,summ);}
}
//1.2
if(i+1<m){if(!temp_vector[i+1][j]){ if(sum_max<(summ)){sum_max=summ;}}}
if(i+1>=m){if(sum_max<(summ)){sum_max=summ;}}
//2
if(i-1>=0 && temp_vector[i-1][j]){
if(test1[i-1][j]){
if(sum_max<(summ)){sum_max=summ;}
}
else{ color1(i-1,j,temp_vector,st,summ);}
}
//2.2
if(i-1>=0){if(!temp_vector[i-1][j]){ if(sum_max<(summ)){sum_max=summ;}}}
if(i-1<0){if(sum_max<(summ)){sum_max=summ;}}
//3
if(j+1<n && temp_vector[i][j+1]){
if(test1[i][j+1]){
if(sum_max<(summ)){sum_max=summ;}
}
else{ color1(i,j+1,temp_vector,st,summ);}}
//3.2
if(j+1<n){if(!temp_vector[i][j+1]){ if(sum_max<(summ)){sum_max=summ;}}}
if(j+1>=n){if(sum_max<(summ)){sum_max=summ;}}
//4
if(j-1>=0 && temp_vector[i][j-1]){
if(test1[i][j-1]){
if(sum_max<(summ)){sum_max=summ;}
}
else{ color1(i,j-1,temp_vector,st,summ);}}
//4.2
if(j-1>=0){if(!temp_vector[i][j-1]){ if(sum_max<(summ)){sum_max=summ;}}}
if(j+1<0){if(sum_max<(summ)){sum_max=summ;}}
}
void color(int i,int j){
test[i][j]=false;
if(i+1<m && test[i+1][j]){
color(i+1,j);}
if(i-1>=0 && test[i-1][j]){
color(i-1,j);
}
if(j+1<n && test[i][j+1]){
color(i,j+1);}
if(j-1>=0 && test[i][j-1]){color(i,j-1);}
}
int main(){
int tc;cin>>tc;
for(int i=0;i<tc;i++){
int mp,np;
cin>>mp;
cin>>np;m=mp;n=np;A.resize(m);test.resize(m);test1.resize(m);int sum=0;
vector<bool> ha1(m,1);
vector<bool> ha2(n,1);
for(int i=0;i<m;i++){A[i].resize(n);test[i].resize(n);test1[i].resize(n);
for(int j=0;j<n;j++){
cin>>A[i][j];sum+=A[i][j];
test[i][j]=true;test1[i][j]=false;
if(A[i][j]==0){test[i][j]=false;ha1[i]=false;ha2[j]=false;}
}
}cout<<endl;
for(int i=0;i<m;i++){cout<<" "<<ha1[i];} cout<<endl;
for(int i=0;i<n;i++){cout<<" "<<ha2[i];} cout<<endl;
cout<<"sum "<<sum<<"\n";
int temp_sum=0;
for(int i=0;i<m;i++){
for(int j=0;j<n;j++){//if(A[i][j]<=8845){cout<<"\nk "<<A[i][j]<<" "<<(8845-A[i][j]);}
if(test[i][j]){
if((i-1)>=0 && test[i-1][j] && (i+1)<m && test[i+1][j] && (j-1)>=0 && test[i][j-1] && (j+1)<n && test[i][j+1] && test[i-1][j-1] && test[i-1][j+1]&& test[i+1][j-1] && test[i+1][j+1]){
temp_sum+=A[i][j];test1[i][j]=true;}
}
// cout<<test1[i][j]<<" ";
}//cout<<"\n";
}
// /*
for(int i=0;i<m;i++){
for(int j=0;j<n;j++){
if(test1[i][j]){if(!((test1[i-1][j]||test1[i+1][j]) && (test1[i][j-1]||test1[i][j+1]))){
temp_sum-=A[i][j]; test1[i][j]=false;}
}
//
// cout<<test1[i][j]<<" ";
}//
// cout<<"\n";
}
// */
//cout<<"\n temp_sum is "<<temp_sum<<endl;
vector<vector<bool> > st(m,vector<bool>(n,0));st=test1;
for(int i=0;i<m;i++){
for(int j=0;j<n;j++){
if(test[i][j] && (!test1[i][j])){
color1(i,j,test,st,0);
}}}
// cout<<"\nsum is "<<(sum_max+temp_sum)<<endl<<endl;
cout<<(sum_max+temp_sum)<<endl;
for(int i=0;i<m;i++){
for(int j=0;j<n;j++){cout<<stamp[i][j]<<" ";} cout<<endl;}
// cout<<max<<endl;
A.clear();
test.clear();
test1.clear();
sum_max=0;
}
cout<<endl;system("pause");
return 0;
}
I'm trying to construct an algorithm that runs at O(nb) time with the following input/question:
input: an array A[1..n] of n different integers and an integer b (i am assuming that the numbers in A are sequential, starting at 1 ending at n, i.e. for n=4 A[1,2,3,4].
question: in how many ways can b be written as the sum of elements of the array when elements in A[] can only be used once?
I've kind of hit a wall on this one. I'm looking for some kind of recursive solution, but I don't see how to avoid using repeat numbers. Like, for instance, if we started at 1 and stored all the ways to make one (just 1) then 2 (just 2) then three (3 or 2+1) etc, it shouldn't be hard to see how many ways we can make larger numbers. But if, for instance, we take 5, we will see that it can be broken into 4+1, and 4 can be further broken down into 3+1, so then we would see 2 solutions (4+1, and 3+1+1), but one of those has a repeat of a number. Am I missing something obvious? Thanks so much!
Recursive and dynamic solutions in C:
#include <stddef.h>
#include <assert.h>
#include <stdio.h>
#include <stdlib.h>
typedef unsigned char uchar;
typedef unsigned int uint;
typedef struct tAddend
{
struct tAddend* pPrev;
uint Value;
} tAddend;
void findRecursiveSolution(uint n, uint maxAddend, tAddend* pPrevAddend)
{
uint i;
for (i = maxAddend; ; i--)
{
if (n == 0)
{
while (pPrevAddend != NULL)
{
printf("+%u", pPrevAddend->Value);
pPrevAddend = pPrevAddend->pPrev;
}
printf("\n");
return;
}
if (n >= i && i > 0)
{
tAddend a;
a.pPrev = pPrevAddend;
a.Value = i;
findRecursiveSolution(n - i, i - 1, &a);
}
if (i <= 1)
{
break;
}
}
}
void printDynamicSolution(uchar** pTable, uint n, uint idx, uint sum, tAddend* pPrevAddend)
{
uchar el = pTable[idx][sum];
assert((el != 0) && (el != 5) && (el != 7));
if (el & 2) // 2,3,6 - other(s)
{
printDynamicSolution(pTable,
n,
idx - 1,
sum,
pPrevAddend);
}
if (el & 4) // self + other(s)
{
tAddend a;
a.pPrev = pPrevAddend;
a.Value = idx + 1;
printDynamicSolution(pTable,
n,
idx - 1,
sum - (idx + 1),
&a);
}
if (el & 1) // self, found a solution
{
tAddend a;
a.pPrev = pPrevAddend;
a.Value = idx + 1;
pPrevAddend = &a;
while (pPrevAddend != NULL)
{
printf("+%u", pPrevAddend->Value);
pPrevAddend = pPrevAddend->pPrev;
}
printf("\n");
}
}
void findDynamicSolution(uint n)
{
uchar** table;
uint i, j;
if (n == 0)
{
return;
}
// Allocate the DP table
table = malloc(sizeof(uchar*) * n);
if (table == NULL)
{
printf("not enough memory\n");
return;
}
for (i = 0; i < n; i++)
{
table[i] = malloc(n + 1);
if (table[i] == NULL)
{
while (i > 0)
{
free(table[--i]);
}
free(table);
printf("not enough memory\n");
return;
}
}
// Fill in the DP table
for (i = 0; i < n; i++)
{
for (j = 0; j <= n; j++)
{
if (i == 0)
{
table[i][j] = (i + 1 == j); // self
}
else
{
table[i][j] = (i + 1 == j) + // self
2 * (table[i - 1][j] != 0) + // other(s)
4 * ((j >= i + 1) && (table[i - 1][j - (i + 1)] != 0)); // self + other(s)
}
}
}
printDynamicSolution(table, n, n - 1, n, NULL);
for (i = 0; i < n; i++)
{
free(table[i]);
}
free(table);
}
int main(int argc, char** argv)
{
uint n;
if (argc != 2 || sscanf(argv[1], "%u", &n) != 1)
{
n = 10;
}
printf("Recursive Solution:\n");
findRecursiveSolution(n, n, NULL);
printf("\nDynamic Solution:\n");
findDynamicSolution(n);
return 0;
}
Output:
for 10:
Recursive Solution:
+10
+1+9
+2+8
+3+7
+1+2+7
+4+6
+1+3+6
+1+4+5
+2+3+5
+1+2+3+4
Dynamic Solution:
+1+2+3+4
+2+3+5
+1+4+5
+1+3+6
+4+6
+1+2+7
+3+7
+2+8
+1+9
+10
See also on ideone.
Let F(x,i) be the number of ways elements of A[1:i] can be summed to get x.
F(x,i+1) = F(x-A[i+1],i) + F(x,i)
That is it!
This is not a dynamic programming solution though. Non-recursive.
Assumption that arr is sorted in your case like [i....j] where a[i] <= a[j]
That's easy enough
void summer(int[] arr, int n , int b)
{
int lowerbound = 0;
int upperbound = n-1;
while (lowerbound < upperbound)
{
if(arr[lowerbound]+arr[upperbound] == b)
{
// print arr[lowerbound] and arr[upperbound]
lowerbound++; upperbound--;
}
else if(arr[lowerbound]+arr[upperbound] < b)
lowerbound++;
else
upperbound--;
}
}
The above program is easily modifiable to a recursive you need to only change the function definition by passing lowerbound and upperbound.
Case for termination is still lowerbound < upperbound
Base case is if arr[lowerbound] +arr[upperbound] == b
Edited based on comments
You will need to use a modified version of integer knapsack problem. The values of [i,j] both need to be modified accordingly. You are having the problem because you are not most probably modifying your i carefully, Increase your i accordingly then their will not be repetition like the one you are having.