Charm Bracelet Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 38909   Accepted: 16862 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 fro…

Charm Bracelet Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 34532   Accepted: 15301 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 fro…

传送门:http://poj.org/problem?id=3624 题目大意:XXX去珠宝店,她需要N件首饰,能带的首饰总重量不超过M,要求不超过M的情况下,使首饰的魔力值(D)最大. 0-1背包入门题. 可构建状态转移方程: dp [ i ] [ v ]= max ( dp[ i-1 ] [ v ], dp[ i-1 ][ v- W[ i ] ]+d[ i ] ] ) 但是这样空间太大,可以用滚动数组解决. for(int i=1;i<=N;i++) { for(int j=M;j>=w[…

题目链接 Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 52318   Accepted: 21912 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…

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 …

题目描述 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),…

题目传送门 解题思路: 一维解01背包,突然发现博客里没有01背包的板子,补上 AC代码: #include<cstdio> #include<iostream> using namespace std; ]; int main() { scanf("%d%d",&n,&m); ;i <= n; i++) { scanf("%d%d",&c,&w); ; j--) if(c <= j) f[j] =…

题目链接:poj 3624 这是最基础的背包问题,特点是:每种物品仅有一件,可以选择放或不放.             用子问题定义状态:即F [i, v]表示前i件物品恰放入一个容量为v 的背包可以             获得的最大价值.则其状态转移方程便是:             F [i, v] = max{F [i − 1, v], F [i − 1, v − Ci ] + Wi }             这个方程非常重要,基本上所有跟背包相关的问题的方程都是由它衍生       …

题目意思: 给出N,M,N表示有N个物品,M表示背包的容量.接着给出每一个物品的体积和价值,求背包可以装在的最大价值. http://poj.org/problem? id=3624 题目分析: o-1背包问题,转化方程.dp[j]:表示容量为j的时候,背包的最大价值 dp[j]=max(dp[j],dp[j-w[i]]+d[i]); AC代码: #include<iostream> #include<cstring> using namespace std; int dp[200…

POJ.3624 Charm Bracelet(DP 01背包) 题意分析 裸01背包 代码总览 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #define nmax 13000 #define nnmax 3500 using namespace std; int dp[nmax]; int w[nnmax],d[nnmax]; int main…