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  • HDU Problem 2062 Bone Collector【01背包】

    Bone Collector

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 51847    Accepted Submission(s): 21829

    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
     
    Sample Output
    14
     
    Author
    Teddy
     
    Source
     
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    状态转移方程:dp[i][j] = max{dp[i+1][j], dp[i+1][j-vol[i]]+val[i]};其中i表示第i个物品,j表示背包当前的容量,而数组中所存放的元素为在当前状态下的最大价值。通过下面的每个状态的最大值向上推。因此,最终的最大状态应该存放在dp[0][v]中。
    #include <cstdio>
    #include <iostream>
    #include <cstring>
    #include <algorithm>
    using namespace std;
    const int MAXN = 1000 + 10;
    const int INF = 0x3f3f3f3f;
    int  n, v, val[MAXN], vol[MAXN], dp[MAXN][MAXN];
    int main() {
        int t; scanf("%d", &t);
        while (t--) {
            scanf("%d %d", &n, &v);
            for (int i = 0; i < n; i++) {
                scanf("%d", &val[i]);
            }
            for (int i = 0; i < n; i++) {
                scanf("%d", &vol[i]);
            }
            memset(dp, 0, sizeof(dp));
            for (int i = n - 1; i >= 0; i--) {
                for (int j = 0; j <= v; j++) {
                    if (j < vol[i]) dp[i][j] = dp[i + 1][j];
                    else dp[i][j] = max(dp[i+1][j], dp[i+1][j-vol[i]]+val[i]);
                }
            }
            printf("%d
    ", dp[0][v]);
        }
        return 0;
    }
     
     


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  • 原文地址:https://www.cnblogs.com/cniwoq/p/6770830.html
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