HDU-2602 Bone Collector 裸01背包

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

題目大意：用有限大小的容器收集總價值盡可能大的骨頭，每個骨頭有自身的價值和占用空間。

分析：簡單的01背包，裸模闆套就行。

AC代碼：

#include <iostream>
#include <cstring>
using namespace std;

int dp[1005][1005],numc[1005],numv[1005];

int main() {
    int T, v, c;
    cin >> T;
    while (T--) {
        cin >> c >> v;
        for (int i = 1; i <= c; i++) {
            cin >> numv[i];
        }
        for (int i = 1; i <= c; i++) {
            cin >> numc[i];
        }
        memset(dp, 0, sizeof(dp));
        for (int i = 1; i <= c; i++) {
            for (int j = 0; j <= v; j++) {
                if (j >= numc[i]) {
                    dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - numc[i]] + numv[i]);
                } else dp[i][j] = dp[i - 1][j];
            }
        }
        cout << dp[c][v] << endl;
    }
}