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## Problema da mochila 0-1 - Paradigmas

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```/**
Notes :
1. 0/1 Knapsack problem
2. u cant take an object more than one time.
suppose, u can stand with maximum 30 kg weight an object has a weight of 10 kg
but u can take the object only once, not 3 times, though u still have 20 kg space
3. calculate all the values up to 30 and save in an array dp
if ur maxWeightLimit = i, thn dp[i] = maxValue u can get
*/
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define INT_MAX 2147483647
#define INT_MIN -2147483648
#define N 1000000
using namespace std;
int main()
{
int testCase;
scanf("%d", &testCase);

while(testCase--) {
int objects;
int values[1000 + 5];
int weight[1000 + 5];
scanf("%d", &objects);

for(int i = 0; i < objects; ++i)
scanf("%d %d", &values[i], &weight[i]);

int dp[30 + 5];
memset(dp, 0, sizeof(dp));

for(int j = 0; j < objects; ++j) {
for(int i = 32; i >= 0; --i) {
if(weight[j] <= i && dp[i] < dp[i - weight[j]] + values[j]) {
dp[i] = dp[i - weight[j]] + values[j];
}
}
}

int g, volume, aux;
scanf("%d", &g);

while(g--) {
scanf("%d", &volume);
printf("Max value = \$%d\n", dp[volume]);
// print solution
aux = volume;

while((aux > 0) && (last_added[aux] != -1)) {
}
}
}

return 0;
}
```