Ok guys heres the problem and its partial soultion, i'd like your help in completing it casue i'm at my wits end.
Problem Statement
Suppose I want to make a decision rationally. One approach is to come up with several categories and weight each category according to its importance. Then I assign scores in each category to the competing alternatives, and pick the alternative with the highest total weighted score. For example, suppose I am buying a new car and I need to decide between a sedan, a minivan, or a sport-utility vehicle (SUV). In consultation with my wife, I come up with four categories
cost (weight 3)
carrying capacity (weight 2)
fuel efficiency (weight 1)
fun (weight 1)
Code:
After studying each vehicle carefully, we give them the following scores:
Vehicle Cost Carrying Capacity Fuel Efficiency Fun Total Score
Sedan 6 3 5 4 6*3+3*2+5*1+4*1 = 33
Minivan 5 5 3 2 5*3+5*2+3*1+2*1 = 30
SUV 4 6 2 6 4*3+6*2+2*1+6*1 = 32
Clearly we should purchase the sedan. Unfortunately, neither of us wants the sedan. I really want the minivan, and my wife really wants the SUV. And so begins the process of rationalization, in which we each try to tweak the numbers to make our choice come out on top. She quickly realizes that by tweaking one number, changing the weight of the fun category from 1 to 2, she can cause the SUV to win with a score of 38 (versus 37 for the sedan and 32 for the minivan). I have to work harder, but if I tweak two numbers, changing the cost score of the minivan from 5 to 6 and the efficiency score of the sedan from 5 to 4, then I can make the minivan win with a score of 33 (versus 32 for both the SUV and the sedan). Note that there are several other tweaks that each of us could have made that would have achieved our respective goals.
Given the initial weights and scores, as well as the zero-based index desired of the alternative that you want to win, determine the minimum number of tweaks needed to make your chosen alternative win. To win, your chosen alternative must end up with a score strictly higher than all the other alternatives--ties are not good enough. A single tweak involves changing the value of a particular weight or a particular score up or down by one. The same number cannot be tweaked more than once, and a tweak may not cause a weight or a score to exceed 9 or drop below 1. If no amount of tweaking can make your chosen alternative win, return -1.
Weights will be given as a int[], and scores will be given as a String[]. Element J of weights is the weight for category J, and element I of scores contains the scores for alternative I. Within element I of scores, character J represents the score for alternative I in category J. In the example above, weights would be { 3, 2, 1, 1 } and scores would be {"6354", "5532", "4626"}, with desired = 2 for the SUV and desired = 1 for the minivan.
Definition
Class: Rationalization
Method: minTweaks
Parameters: int[], String[], int
Returns: int
Method signature: int minTweaks(int[] weights, String[] scores, int desired)
Constraints
- Weights contains between 2 and 10 elements, inclusive.
- Each element of weights is between 1 and 9, inclusive.
- Scores contains between 2 and 10 elements, inclusive.
- Each element of scores contains exactly W characters, where W is the number of elements in weights.
- Each character in scores is a digit between '1' and '9', inclusive.
- Desired is between 0 and S-1, inclusive, where S is the number of elements in scores.
Examples
1) {3, 2, 1, 1}
{"6354", "5532", "4626"}
2
Returns: 1
The example above was trying to make the SUV win.
2) {3, 2, 1, 1}
{"6354", "5532", "4626"}
1
Returns: 2
The example above was trying to make the minivan win.
3) {3, 2, 1}
{"555", "333"}
1
Returns: -1
Option 1 can never beat option 0. The best it can do is tie.
4) {1, 2, 3, 3}
{"9234", "1334"}
1
Returns: 3
Unfortunately, we can't drop the weight of the first category to 0.
5) {8, 2}
{"55", "92"}
0
Returns: 6
6) {2, 8, 7, 3, 6, 5, 2, 4, 7, 2}
{"9197287893", "9492555365", "3459972761", "4886112198", "5963616776",
"6325897129", "3248793133", "7984474438", "4518544769", "1592681682"}
5
Returns: 17
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GENERAL ALGORITHM
this is not an efficient solution but it will work
as mentioned in the problem, many rationalization can be there and this short time i managed to find this simple and just working form
1. get weight in an array of integer W[]
check for valid values(1 to 9)
check for no of elements(2 to 10)
2. get scores in a string array S[]
get limit of array(2 to 10)
enter the srings one by one checking each string for-
lingth of string = size of weight array
every character should be a digit between 1 to 9
set n = size of score array
3. get desire (0 to n)
let option be x
4. caldulate total score
create an integer array A[] of size same as n
create a temporary array T[]of integers with size same as weight array
repeat for every string of score array
store every string character of a single string as integer value in T[] with first character at first location and so on.
perform W[i]*T[i] and store it in A[i] for every i
select largest integer in array A and get its location
if Amax is a location x
return 0
exit
else
move to next step
5 Rationalization
Set tweak=0
find D=Amax-A[x]
store S[x] in T[] as integer numbers
arrange all the indices of s[] in a temporary array in descending order of their content
repeat for all i in the temporary array
if S[i]+1 gives total of A[x]>D
tweak+=1
elseif S[i]+1,s[i-1]+1 gives total of A[x]>D
tweak+=2
-
-
elseif s[i]+...,s[i-upperbound]+1 gives total of A[x]>D
tweak+=9
else
tweak=-1
if tweak = -1 then
{arrange all the indices of W[] in a temporary array in descending order of their content
repeat for all i in the temporary array
if W[i]>D
tweak+=1
elseif W[i]+W[i-1]>D
tweak+=2
elseif W[i]+W[i-1]+W[i-2]>D
tweak+=3
-
-
elseif W[i]+...+W[i-upperbound]>D
tweak+=9
else
tweak=-1
}
end if
return tweak
6 End
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PART of the PROGRAM (its not fully based on above algorithm)
STEP 1 & 2. If u tweak weight say ‘j’, then gain made by the ‘desired’ total score is equal to the difference between score ‘j’ of ‘desired’ and score ‘j’ of ‘winner’ ( multiplied by amount of tweak, in this case +/-1 ).
STEP 3 & 4. If u tweak score say ‘j’ of ‘desired’, then the gain made by the ‘desired’ total score is equal to the weight ‘j’ (multiplied by amount of tweak, in this case +/-1 ).
Code:
#include <conio.h>
#include <iostream.h>
#include <io.h>
int t=0; //no. of tweaks
int w[4]; //weights
int s[2][4]; //scores
int tot[2]; //total scores
int d; //desired element
int g[2]; //points losing by 'd' to win
int wintot=-1; //winner total
int win=-1; //winner element
void main() {
//compute total scores
computetotal();
if(d==win) cout << t << " Tweaks"; exit(0);
//tweak weights +ve changes
int tflag=0;
for(int j=0;j<4;j++) {
tflag=0;
g[win]=s[d][j]-s[win][j];
if(g[win]>0) {
for(int i=0;i<2;i++) {
g[i]=s[d][j]-s[i][j];
if(g[i]<=tot[d]-tot[i])
tflag=1;
}
if(!tflag) {
w[j]++; //tweak weight j
t++;
}
computetotal();
if(d==win) cout << t << " Tweaks"; exit(0);
else
//tweak weights -ve changes
{int tflag=0;
for(int j=0;j<4;j++) {
tflag=0;
g[win]=s[win][j]-s[d][j];
if(g[win]>0) {
for(int i=0;i<2;i++) {
g[i]=s[i][j]-s[d][j];
if(g[i]<=tot[d]-tot[i])
tflag=1;
}
if(!tflag) {
w[j]++; //tweak weight j
t++;
}
computetotal();
if(d==win) cout << t << " Tweaks"; exit(0);}
}
}
}
//tweak scores +ve changes
//tweak scores -ve changes
}
void computetotal() {
for(int i=0;i<2;i++) {
for(int j=0;j<4;j++)
tot[i]+=s[i][j]*w[i];
if(tot[i]>wintot) {
wintot=tot[i];
win=i;
}
}
}
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I would like ur help in implementing the score section of the program.
Help please,
Boltress
Linear Mode
