POJ3624 Charm Bracelet 【01背包】

最新推荐文章于 2019-08-06 15:11:09 发布

转载最新推荐文章于 2019-08-06 15:11:09 发布 · 175 阅读

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原文链接：http://www.cnblogs.com/zhchoutai/p/8452787.html

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#数据结构与算法

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 weightW_i (1 ≤ W_i ≤ 400), a 'desirability' factor D_i (1 ≤ D_i ≤ 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: W_i and D_i

Output

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

Sample Input

Sample Output

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\n", 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\n", 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1288234044/1024=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\n", dp[n][m]);
    return 0;
}

转载于:https://www.cnblogs.com/zhchoutai/p/8452787.html