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  • POJ3624 Charm Bracelet 【01背包】

    Charm Bracelet
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 22621   Accepted: 10157

    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 weightWi (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

    题意:给定物品数量n和背包容量m,n个物品的重量weight和价值val,求能获得的最大价值。

    题解:状态转移方程:dp[i][j] = max(dp[i-1][j], dp[i-1][j-w[i]] + v[i]);当中状态dp[i][j]表示前i个物品放在容量为j的背包中能获得的最大价值。

    二维数组能够压缩成一维以节省空间,可是内层循环须要倒序。

    原始版本号:耗时360ms

    #include <stdio.h>
    #define maxn 12882
    
    int dp[maxn];
    
    int max(int a, int b){ return a > b ? a : b; }
    
    int main()
    {
        int n, totalWeight, i, j, weight, val;
        scanf("%d%d", &n, &totalWeight);
        for(i = 1; i <= n; ++i){
            scanf("%d%d", &weight, &val);
            for(j = totalWeight; j; --j){
                if(j >= weight) dp[j] = max(dp[j], dp[j - weight] + val);
            }
        }
        printf("%d
    ", dp[totalWeight]);
        return 0;
    }<span style="font-family:FangSong_GB2312;">
    </span>

    优化后的代码:耗时219ms

    #include <stdio.h>
    #define maxn 12882
    
    int dp[maxn];
    
    int main()
    {
        int n, totalWeight, i, j, weight, val;
        scanf("%d%d", &n, &totalWeight);
        for(i = 1; i <= n; ++i){
            scanf("%d%d", &weight, &val);
            for(j = totalWeight; j; --j){
                if(j >= weight && dp[j - weight] + val > dp[j]) 
    				dp[j] = dp[j - weight] + val;
            }
        }
        printf("%d
    ", dp[totalWeight]);
        return 0;
    }
    

    用二维dp数组写了一个,果断的超了内存,占用内存大概12882*3404*4/1024=171兆。题目限制是65兆

    TLE:

    #include <stdio.h>
    #define maxn 12882
    
    int dp[3404][maxn];
    
    int max(int a, int b){ return a > b ? a : b; }
    
    int main()
    {
        int n, m, weight, val, i, j;
        scanf("%d%d", &n, &m);
        for(i = 1; i <= n; ++i){
            scanf("%d%d", &weight, &val);
            for(j = 1; j <= m; ++j)
                if(j >= weight)
                    dp[i][j] = max(dp[i-1][j], dp[i-1][j-weight] + val);
                else dp[i][j] = dp[i-1][j];
        }
        printf("%d
    ", dp[n][m]);
        return 0;
    }
    



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  • 原文地址:https://www.cnblogs.com/zhchoutai/p/8452787.html
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