| 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\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\n", dp[n][m]);
return 0;
}