P2871 [USACO07DEC]手鏈Charm Bracelet
題目描述
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.
有N件物品和一個容量為V的背包。第i件物品的重量是c[i],價值是w[i]。求解將哪些物品裝入背包可使這些物品的重量總和不超過背包容量,且價值總和最大。
輸入輸出格式
輸入格式:
-
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
第一行:物品個數N和背包大小M
第二行至第N+1行:第i個物品的重量C[i]和價值W[i]
輸出格式:
- Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints
輸出一行最大價值。
輸入輸出樣例
輸入樣例#1: 復制4 6 1 4 2 6 3 12 2 7輸出樣例#1: 復制
23
//二維DP會 炸 //WA 80 #include<bits/stdc++.h> #defineView CodeN 12881 using namespace std; int n,m,w[N],v[N],dp[N]; int main() { scanf("%d%d",&n,&m); for(int i=1;i<=n;i++) scanf("%d%d",&w[i],&v[i]); for(int i=1;i<=n;i++) for(int j=0;j<=m;j++) { if(j<w[i]) dp[i][j]=dp[i-1][j]; else dp[i][j]=max(dp[i-1][j],dp[i-1][j-w[i]]+v[i]); } cout<<dp[n][m]; }
//一維DP AC #include<bits/stdc++.h> #define N 50000 using namespace std; int n,m,w[N],v[N],dp[N]; int main() { scanf("%d%d",&n,&m); for(int i=1;i<=n;i++) scanf("%d%d",&w[i],&v[i]); for(int i=1;i<=n;i++) for(int j=m;j>0;j--) { if(j>=w[i]) dp[j]=max(dp[j],dp[j-w[i]]+v[i]); } printf("%d",dp[m]); }View Code
//01背包板子題
P2871 [USACO07DEC]手鏈Charm Bracelet