Bone Collector
Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave …
The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ?
Input
The first line contain a integer T , the number of cases.
Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.
Output
One integer per line representing the maximum of the total value (this number will be less than 2 31).
Sample Input
1
5 10
1 2 3 4 5
5 4 3 2 1
Sample Output
14
一個(gè)簡單背包, 根據(jù)狀態(tài)轉(zhuǎn)移方程就能做了:
dp[j] = max(dp[j], dp[j-w[i]] + v[i]);
#include <bits/stdc++.h>
using namespace std;
int main() {
int T;
int n, v;
int value[10005];
int weight[10005];
scanf("%d", &T);
while(T--) {
int dp[10005];
memset(dp, 0, sizeof dp);
scanf("%d%d", &n, &v);
for(int i = 0; i < n; i++) {
scanf("%d", &value[i]);
}
for(int j = 0; j < n; j++) {
scanf("%d", &weight[j]);
}
for(int i = 0; i < n; i++) {
for (int j = v; j >= weight[i]; j--) {
dp[j] = max(dp[j], dp[j - weight[i]] + value[i]);
}
}
printf("%d\n", dp[v]);
}
return 0;
}
