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
Source
dp[i][j]代表当背包体积为j时，前i件物品可以达到的最大价值。
当i为0时，前0件物品价值只能是0。所以第一行全是0；当第i件物品重量大于j时，dp[i][j] = dp[i - 1][j]，否则dp[i][j] = max(dp[i - 1][j],
dp[i - 1][j - w[i] ] + v[i])。
为了优化内存，可以使用一维数组，但是j要从M递减到1，否则从1到M的话前面的会覆盖掉后面要用到的值。
1 #include <iostream> 2 #include <vector> 3 #include <algorithm> 4 using namespace std; 5 6 vector<int> dp; 7 8 int main() { 9 int N, M, i, j; 10 cin >> N >> M; 11 vector<int> W(N + 1), D(N + 1); 12 for (i = 1; i <= N; i++) 13 cin >> W[i] >> D[i]; 14 dp.resize(M + 1); 15 for (i = 1; i <= N; i++) { 16 for (j = M; j >= 1; j--) { 17 if (W[i] <= j) 18 dp[j] = max(dp[j], dp[j - W[i]] + D[i]); 19 } 20 } 21 cout << dp[M] << endl; 22 return 0; 23 }