Problem
You own a milkshake shop. There are N different flavors that you can prepare, and each flavor can be prepared "malted" or "unmalted". So, you can make 2N different types of milkshakes.
Each of your customers has a set of milkshake types that they like, and they will be satisfied if you have at least one of those types prepared. At most one of the types a customer likes will be a "malted" flavor.
You want to make N batches of milkshakes, so that:
- There is exactly one batch for each flavor of milkshake, and it is either malted or unmalted.
- For each customer, you make at least one milkshake type that they like.
- The minimum possible number of batches are malted.
Find whether it is possible to satisfy all your customers given these constraints, and if it is, what milkshake types you should make.
If it is possible to satisfy all your customers, there will be only one answer which minimizes the number of malted batches.
Input
- One line containing an integer C, the number of test cases in the input file.
For each test case, there will be:
- One line containing the integer N, the number of milkshake flavors.
- One line containing the integer M, the number of customers.
- M lines, one for each customer, each containing:
- An integer T >= 1, the number of milkshake types the customer likes, followed by
- T pairs of integers "X Y", one for each type the customer likes, where X is the milkshake flavor between 1 and N inclusive, and Y is either 0 to indicate unmalted, or 1 to indicated malted. Note that:
- No pair will occur more than once for a single customer.
- Each customer will have at least one flavor that they like (T >= 1).
- Each customer will like at most one malted flavor. (At most one pair for each customer has Y = 1).
All of these numbers are separated by single spaces.
Output
- C lines, one for each test case in the order they occur in the input file, each containing the string "Case #X: " where X is the number of the test case, starting from 1, followed by:
- The string "IMPOSSIBLE", if the customers' preferences cannot be satisfied; OR
- N space-separated integers, one for each flavor from 1 to N, which are 0 if the corresponding flavor should be prepared unmalted, and 1 if it should be malted.
Limits
Small dataset
C = 100
1 <= N <= 10
1 <= M <= 100
Large dataset
C = 5
1 <= N <= 2000
1 <= M <= 2000
The sum of all the T values for the customers in a test case will not exceed 3000.
Sample
Input
|
Output
|
2
5
3
1 1 1
2 1 0 2 0
1 5 0
1
2
1 1 0
1 1 1
|
Case #1: 1 0 0 0 0
Case #2: IMPOSSIBLE
|
In the first case, you must make flavor #1 malted, to satisfy the first customer. Every other flavor can be unmalted. The second customer is satisfied by getting flavor #2 unmalted, and the third customer is satisfied by getting flavor #5 unmalted.
In the second case, there is only one flavor. One of your customers wants it malted and one wants it unmalted. You cannot satisfy them both.
Analysis
On the surface, this problem appears to require solving the classic problem "Satisfiability," the canonical example of an NP-complete problem. The customers represent clauses, the milkshake flavors represent variables, and malted and unmalted flavors represent whether the variable is negated.
We are not evil enough to have chosen a problem that hard! The restriction that makes this problem easier is that the customers can only like at most one malted flavor (or equivalently, the clauses can only have at most one negated variable.)
Using the following steps, we can quickly find whether a solution exists, and if so, what the solution is.
- Start with every flavor unmalted and consider the customers one by one.
- If there is an unsatisfied customer who only likes unmalted flavors, and all those flavors have been made malted, then no solution is possible.
- If there is an unsatisfied customer who has one favorite malted flavor, then we must make that flavor malted. We do this, then go back to step 2.
- If there are no unsatisfied customers, then we already have a valid solution and can leave the remaining flavors unmalted.
Notice that whenever we made a flavor malted, we were forced to do so. Therefore, the solution we got must have the minimum possible number of malted flavors.
With clever data structures, the above algorithm can be implemented to run in linear time.
More information:
The Satisfiability problem - Horn clauses
Source Code
#include <iostream>
using namespace std;
#define Rep(i,n) for (int i(0),_n(n); i<_n; ++i)
struct Flavor
{
int X;
char Y;
};
struct Customer{
int T;
Flavor* F;
Customer() {
F = NULL;
}
~Customer() {
if(NULL!=F) {
delete[] F;
F = NULL;
}
}
void Init(int t) {
T = t;
F = new Flavor[T];
}
void SetFlavor(int i, int X, int Y) {
F[i].X = X;
F[i].Y = Y;
}
int GetFlavorX(int i) {
return F[i].X;
}
int GetFlavorY(int i) {
return F[i].Y;
}
bool IsSatisfied() {
return T==0;
}
void Satisfy() {
T=0;
if(NULL!=F) {
delete[] F;
F = NULL;
}
}
bool IsSatisfing(int i, int *f) {
return f[F[i].X]==F[i].Y;
}
void SetMalted(int i, int *f) {
f[F[i].X] = 1;
}
};
int main()
{
int C;
FILE *fp = fopen("A.out", "w");
scanf("%d", &C);
Rep(c, C) {
int N;
scanf("%d", &N);
int* f = new int[N+1];
Rep(i ,N+1) {
f[i]=0;
}
int M;
scanf("%d", &M);
Customer* customer = new Customer[M];
Rep(m, M) {
int T;
scanf("%d", &T);
customer[m].Init(T);
Rep(t, T) {
int X, Y;
scanf("%d%d", &X,&Y);
customer[m].SetFlavor(t,X,Y);
}
}
bool findSolution = true;
int m = 0;
while(m<M) {
if(customer[m].IsSatisfied()) {
m++;
continue;
}
bool malted = false;
int idx;
bool satisfied = false;
Rep(t, customer[m].T) {
if(customer[m].GetFlavorY(t)==1) {
malted = true;
idx = t;
}
if(customer[m].IsSatisfing(t,f)) {
satisfied = true;
}
}
if(!satisfied) {
if(malted) {
customer[m].SetMalted(idx,f);
customer[m].Satisfy();
m=0;
} else {
findSolution = false;
break;
}
} else {
m++;
}
}
fprintf(fp,"Case #%d: ", c+1);
if(findSolution) {
Rep(i ,N) {
fprintf(fp,"%d ", f[i+1]);
}
fprintf(fp,"\n");
} else {
fprintf(fp,"IMPOSSIBLE\n");
}
delete[] customer;
delete[] f;
}
fclose(fp);
}