• POJ3624 Charm Bracelet(典型01背包问题)


    Time Limit: 1000MS          Memory Limit: 65536K          Total Submissions: 32897          Accepted: 14587

    Description

    Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course, she'd like to fill it with the best charms possible from the N (1 ≤ N ≤ 3,402) available charms. Each charm i in the supplied list has a weight Wi (1 ≤ Wi ≤ 400), a 'desirability' factor Di (1 ≤ Di ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more than M (1 ≤ M ≤ 12,880).

    Given that weight limit as a constraint and a list of the charms with their weights and desirability rating, deduce the maximum possible sum of ratings.

    Input

    * Line 1: Two space-separated integers: N and M
    * Lines 2..N+1: Line i+1 describes charm i with two space-separated integers: Wi and Di

    Output

    * Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints

    Sample Input

    4 6
    1 4
    2 6
    3 12
    2 7

    Sample Output

    23

    一、题目大意

          有N个物品,分别有不同的重量Wi和价值Di,Bessie只能带走重量不超过M的物品,要是总价值最大,并输出总价值。

     

    二、解题思路

         典型的动态规划题目,用一个数组记录背包各个重量的最优解,不断地更新直到穷尽所有可能性。

         状态更新公式:state[weight] = max{state[weight-W[i]]+D[i], state[weight]}

     

    三、具体代码 

     1 // 背包问题(动态规划)
     2 #include <iostream> 
     3 #include <cstdio>
     4 #include <cstring>
     5 #define MAXN 3402
     6 #define MAXM 12880
     7 using namespace std;
     8 
     9 int main(){
    10     int N, M, W[MAXN+5], D[MAXN+5], dp[MAXM+5];
    11     while(scanf("%d%d", &N, &M) != EOF){
    12         for(int i=0; i<N; i++){
    13             scanf("%d%d", &W[i], &D[i]);
    14         }
    15         memset(dp, 0, sizeof(dp));
    16         for(int i=0; i<N; i++){
    17             for(int left_w=M; left_w>=W[i]; left_w--){
    18                 dp[left_w] = max(dp[left_w-W[i]]+D[i], dp[left_w]);
    19             }
    20         }
    21         printf("%d\n", dp[M]);
    22     }
    23     
    24     return 0;
    25 } 
    View Code
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  • 原文地址:https://www.cnblogs.com/jinglecjy/p/5674796.html
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