标签：int put describe return miss desc hat integer break
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 47440 | Accepted: 20178 |
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
如果用dp[n][m]：前n个物品不超过体积m的最大价值，状态转移方程为：
dp[i][j] = 0(i==0orj==0)//没有物品，背包体积再大的最大价值也是0；再多物品，背包体积为0都塞不下。
dp[i][j]=dp[i-1][j](j<d[i])//当前背包塞不下第i个物品，最大价值和第i-1个物品不大于体积j的最大价值是一样的。
dp[i][j]=max(dp[i-1][j],dp[i-1][j-d[i]]+w[i])//当前背包若能装下第i个物品，那么就在装第i个物品后的最大价值和不装第i个物品的最大价值中取最大值
但由于N和M的范围太大，开二维数组会爆内存，分析求解过程发现求解dp[i][j]时只与它上一行的正上方和上一行左边的某个值有关，也就是只用到了它上面那行，可以用一个滚动的一维数组求解，把新值放到上面的位置，但要注意顺序必须是从右到左，不然会覆盖掉有用的值。
1 #include<iostream> 2 #include<cstring> 3 #include<algorithm> 4 using namespace std; 5 int main() 6 { 7 int n, m; 8 int d[13010], w[13010]; 9 int dp[13010];//因为求解该行时只与上一行有关，所以可以用滚动数组，dp[j]表示前i个物品在不超过体积j的最大价值 10 cin >> n >> m; 11 for (int i = 1; i <= n; ++i) 12 { 13 cin >> d[i] >> w[i]; 14 } 15 memset(dp, 0, sizeof(dp)); 16 for (int i = 1; i <= n; ++i) 17 { 18 for (int j = m; j >= 1; --j)//从右往左求解，把新求出的值保存在上面的位置，因为被覆盖的值只与它下面和下一行的右边有关 19 { 20 if (j >= d[i]) 21 dp[j] = max(dp[j], dp[j - d[i]] + w[i]); 22 else //如果j-d[i]<0，说明当前背包的容量不够放第i个物品，那放前i个物品的最大价值与放前i-1个物品的最大价值是相等的， 23 break;//又因为从右往左求值，左边的j只会更小，因此可跳过节省时间 24 25 } 26 } 27 cout << dp[m] << endl; 28 return 0; 29 }
标签：int put describe return miss desc hat integer break
原文地址：https://www.cnblogs.com/knmxx/p/9531340.html
