POJ3624 Charm Bracelet 【01背包】

Charm Bracelet

Time Limit: 1000MS Memory Limit: 65536KTotal 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

4 61 42 63 122 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]);其中状态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。二维数组可以压缩成一维以节省空间，但是内层循环需要倒序。

原始版本：耗时360ms

#include <stdio.h>#define maxn 12882int 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 12882int 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12882*3404*4/1024=171兆，题目限制是65兆

TLE：

#include <stdio.h>#define maxn 12882int 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;}