OJ Board

Perhaps // before @end_of_source_code is giving trouble.

Remove // from @begin_of_source_code also.

Take care to put both on independant lines.

[cpp]
/* @JUDGE_ID: 33655HX 102 C++ */
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;

#ifndef _BINPACKING_H_
#define _BINPACKING_H_
class BinPacking
{
private:
vector<int> color;
vector<int> input_data;
int min_number;
public:
BinPacking();
~BinPacking();
void Init();
void Input(int);
void Output();
void Evaluate_Minimum_Moving();
};

BinPacking::BinPacking()
{
Init();//initialize
}

BinPacking::~BinPacking()
{
color.clear();
input_data.clear();
}

void BinPacking::Init()
{
color.clear();
input_data.clear();
color.push_back(0);
color.push_back(1);
color.push_back(2);
min_number=INT_MIN;

}

void BinPacking::Input(int num)
{
input_data.push_back(num);
}

void BinPacking::Output()
{
vector<int>::iterator vi;
for(vi=color.begin(); vi!=color.end(); vi++) {
switch(*vi) {
case 0:
cout << "B";
break;
case 1:
cout << "G";
break;
case 2:
cout << "C";
break;

default:
break;
}
}
cout << " ";
cout << min_number;
}

void BinPacking::Evaluate_Minimum_Moving()
{
int tmp1, tmp2, tmp3;
int min=INT_MAX;
vector<int> tmp_v;
tmp_v.assign(color.begin(), color.end());
do {
tmp1=tmp2=tmp3=0;
//bin 1
tmp1=input_data[3+tmp_v[0]]+input_data[6+tmp_v[0]];

//bin 2
tmp2 = input_data[tmp_v[1]]+input_data[tmp_v[1]+6];

//bin 3
tmp3 = input_data[tmp_v[2]] + input_data[3+tmp_v[2]];

if(min>tmp1+tmp2+tmp3) {
min=tmp1+tmp2+tmp3;
color.assign(tmp_v.begin(), tmp_v.end());
}
} while(next_permutation(tmp_v.begin(), tmp_v.end()));
min_number=min;
}

#endif

int main()
{
int tmp, cnt;
BinPacking bpobj;
while(true) {
cnt=0;
for(int i=0; i<9; i++) {
cin>>tmp;
if(cin.eof()) exit(0);
bpobj.Input(tmp);
}
bpobj.Evaluate_Minimum_Moving();
bpobj.Output();
bpobj.Init();
cout << endl;
}
return 0;
}
[/cpp]

102 why WA?

I didn't see all the code.

Here is input:

1 2 3 4 5 6 7 8 9

It should be:

BCG 30

But your code appears:

BGC 30

Give you a advice:

Test all the test case the problem before you sumbit.

I have some test case if you need you may ask me.

BGC 30
bin 1 : 11
bin 2 : 10
bin 3 : 9

BCG 30
bin 1 : 11
bin 2 : 12
bin 3 : 7

------------------

therefore, are both 'BGC 30' and 'BCG 30' all right?

If more than one order of brown, green, and clear bins yields the minimum number of movements then the alphabetically first string representing a minimal configuration should be printed.

hope this help.

^^*

This is a very simple problem

you dont need to make a learge code like you give.

you will always get three glasses and so yo have only 6 combination

you need to check that combination in lexographicaly. i.e.,

BCG
BGC
CBG
CGB
GBC
GCB

my code worked for the sample input that was in the problem, but when I submitted it, i received a WA. Somebody Please help!
here is my code:
[cpp]#include <iostream.h>

struct Bin{
int brown;
int green;
int clear;
};

Bin bin[3];

inline int Calculate_Cost(char a[]);
inline bool less(char a[], char r[]);
void Perm(char a[], int k, int n,int& cost,char r[]);

void main(){
int num;
while(cin>>num){
bin[0].brown=num;
cin>>num;
bin[0].green=num;
cin>>num;
bin[0].clear=num;
cin>>num;
bin[1].brown=num;
cin>>num;
bin[1].green=num;
cin>>num;
bin[1].clear=num;
cin>>num;
bin[2].brown=num;
cin>>num;
bin[2].green=num;
cin>>num;
bin[2].clear=num;
char ch[3];
char result[3];
ch[0]='B';ch[1]='C';ch[2]='G';
result[0]='B';result[1]='C';result[2]='G';
int min_cost=32767;
Perm(ch,0,3,min_cost,result);
for (int i=0; i<3; i++) cout << result;
cout <<' '<<min_cost<< endl;
}
}

void Perm(char a[], int k, int n,int& cost,char r[])
{
if (k==n) { // Output permutation.
int x,i;
for ( i=0;i<3;i++) cout<<a<<' ';
cout<<endl;
x=Calculate_Cost(a);
if(x<cost || (x==cost && less(a,r))){
cost=x;
for ( i=0;i<3;i++) r=a;
}
}
else // a[k:n] has more than one permutation.
// Generate these recursively.
for (int i=k; i<n; i++) {
char t=a[k]; a[k]=a; a=t;
Perm(a, k+1, n,cost,r);
// All permutations of a[k+1:n]
t=a[k]; a[k]=a; a=t;
}
}

inline int Calculate_Cost(char a[]){
int y=0;
switch(a[0]){
case 'B': y=y+bin[0].green+bin[0].clear;
break;
case 'G': y=y+bin[0].brown+bin[0].clear;
break;
case 'C': y=y+bin[0].brown+bin[0].green;
}
switch(a[1]){
case 'B': y=y+bin[1].green+bin[1].clear;
break;
case 'G': y=y+bin[1].brown+bin[1].clear;
break;
case 'C': y=y+bin[1].brown+bin[1].green;
}
switch(a[2]){
case 'B': y=y+bin[2].green+bin[2].clear;
break;
case 'G': y=y+bin[2].brown+bin[2].clear;
break;
case 'C': y=y+bin[2].brown+bin[2].green;
}
return y;
}

inline bool less(char a[], char r[]){
for (int i=0;i<3;i++){
if ((int)a>(int)r)
return false;
}
return true;
}[/cpp]
thnx,
Joe

For the purposes of this problem, each bin has infinite capacity and the only constraint is moving the bottles so that each bin contains bottles of a single color. The total number of bottles will never exceed 2^31.

The data type int cannot accomodate the input limit. I suggest you to change int to long int.

Most probably your mistake is here:

Really??
I thought int type can only accomodate 32767 for all the compiler???
Now that's something new!!!

Yep, int is 32-bits signed integer since 64-bit compilers were here years ago ...

I noticed that the univ I graduated from still uses old compilers like 16-bits BC ... Even their Intel-Assembly language programming is still stuck with 16-bits programming when the processor has evolved a great deal as of today *sigh* ...

-turuthok-

turuthok wrote:Yep, int is 32-bits signed integer since 64-bit compilers were here years ago ...

I noticed that the univ I graduated from still uses old compilers like 16-bits BC ... Even their Intel-Assembly language programming is still stuck with 16-bits programming when the processor has evolved a great deal as of today *sigh* ...

-turuthok-

I don't know why universities don't include new apparatus and subject materials in the course

I am with the same voice with you and thanks for you comment

hi,
I tried to change all int's to long but still I got WA
here is my code.
[cpp]#include <iostream.h>
#include <math.h>

struct Bin{
long brown;
long green;
long clear;
};

Bin bin[3];

inline long Calculate_Cost(char a[]);
inline bool less(char a[], char r[]);
void Perm(char a[], int k, int n,long& cost,char r[]);

void main(){
long num;
while(cin>>num){
bin[0].brown=num;
cin>>num;
bin[0].green=num;
cin>>num;
bin[0].clear=num;
cin>>num;
bin[1].brown=num;
cin>>num;
bin[1].green=num;
cin>>num;
bin[1].clear=num;
cin>>num;
bin[2].brown=num;
cin>>num;
bin[2].green=num;
cin>>num;
bin[2].clear=num;
char ch[3];
char result[3];
ch[0]='B';ch[1]='C';ch[2]='G';
result[0]='B';result[1]='C';result[2]='G';
long min_cost=32767;
Perm(ch,0,3,min_cost,result);
for (int i=0; i<3; i++) cout << result;
cout <<' '<<min_cost<< endl;
}
}

void Perm(char a[], int k, int n,long& cost,char r[])
{
if (k==n) { // Output permutation.
int x,i;
x=Calculate_Cost(a);
if(x<cost || (x==cost && less(a,r))){
cost=x;
for ( i=0;i<3;i++) r=a;
}
}
else // a[k:n] has more than one permutation.
// Generate these recursively.
for (int i=k; i<n; i++) {
char t=a[k]; a[k]=a; a=t;
Perm(a, k+1, n,cost,r);
// All permutations of a[k+1:n]
t=a[k]; a[k]=a; a=t;
}
}

inline long Calculate_Cost(char a[]){
int y=0;
switch(a[0]){
case 'B': y=y+bin[0].green+bin[0].clear;
break;
case 'G': y=y+bin[0].brown+bin[0].clear;
break;
case 'C': y=y+bin[0].brown+bin[0].green;
}
switch(a[1]){
case 'B': y=y+bin[1].green+bin[1].clear;
break;
case 'G': y=y+bin[1].brown+bin[1].clear;
break;
case 'C': y=y+bin[1].brown+bin[1].green;
}
switch(a[2]){
case 'B': y=y+bin[2].green+bin[2].clear;
break;
case 'G': y=y+bin[2].brown+bin[2].clear;
break;
case 'C': y=y+bin[2].brown+bin[2].green;
}
return y;
}

inline bool less(char a[], char r[]){
for (int i=0;i<3;i++){
if ((int)a>(int)r)
return false;
}
return true;
}[/cpp]
Joe

Did you read my previous post??? Pay attention man: Although you changed int to long (and you didn't have to do it!!) you did NOT change the min_cost value -> initialize it to 2^31-1 or 2^32-1 or something of that magnitude... 32767 is obviously too small!

Ivor