HDU2602:Bone Collector(01揹包)

Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 93341 Accepted Submission(s): 38176

Problem Description
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 231).

Sample Input
1
5 10
1 2 3 4 5
5 4 3 2 1

01揹包公式題，可利用滾動數組來優化空間：）。

下面附上ac代碼：

#include <bits/stdc++.h>
#include <iostream>
#include <string.h>
#include <algorithm>
#include <ctype.h>
#include <cstdlib>
#include <cmath>
#include <queue>
#include <stdlib.h>
#include <time.h>
using namespace std;
#define INF 0x7ffffff
#define mod 10000019
typedef long long int ll;
struct node {
    int vlv, vol;
}record[2000];
ll dp[1100];
ll t, n, v;
int main() {
    cin >> t;
    while(t--) {
        cin >> n >> v;
        for(int i = 1; i <= n; i++) {
            cin >> record[i].vlv;
        }
        for(int i = 1; i <= n; i++) {
            cin >> record[i].vol;
        }
        memset(dp, 0, sizeof(dp));
        for(int i = 1; i <= n; i++) {
            for(int j = v; j >= record[i].vol; j--) { //滾動數組
                dp[j] = max(dp[j], dp[j - record[i].vol] + record[i].vlv);
            }
        }
        cout << dp[v] << endl;
    }
    return 0;
}