POJ3624--01背包】的更多相关文章

Charm Bracelet 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 fro…
Charm Bracelet Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 30417   Accepted: 13576 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…
转载请注明出处:http://blog.csdn.net/u012860063 贴一个自觉得解说不错的链接:http://www.cppblog.com/tanky-woo/archive/2010/07/31/121803.html 模版就直接贴代码: 01背包模板: /* 01背包问题 01背包问题的特点是,">每种物品仅有一件.能够选择放或不放. 01背包问题描写叙述: 有N件物品和一个容量为V的背包. 第i件物品的重量是c[i],价值是w[i]. 求解将哪些物品装入背包可使这些物品…
题意:过山车有n个区域,一个人有两个值F,D,在每个区域有两种选择: 1.睁眼: F += f[i], D += d[i] 2.闭眼: F = F ,     D -= K 问在D小于等于一定限度的时候最大的F. 解法: 用DP来做,如果定义dp[i][j]为前 i 个,D值为j的情况下最大的F的话,由于D值可能会增加到很大,所以是存不下的,又因为F每次最多增加20,那么1000次最多增加20000,所以开dp[1000][20000],dp[i][j]表示前 i 个,F值为j的情况下最小的D.…
Team Them Up! Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 7608   Accepted: 2041   Special Judge Description Your task is to divide a number of persons into two teams, in such a way, that: everyone belongs to one of the teams; every t…
传送门 Description Hasan and Bahosain want to buy a new video game, they want to share the expenses. Hasan has a set of N coins and Bahosain has a set of M coins. The video game costs W JDs. Find the number of ways in which they can pay exactly W JDs su…
题目链接: http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1085 题意: 中文题诶~ 思路: 01背包模板题. 用dp[i][j]表示到第i个物品花去j空间能存储的最大价值, 那么很显然有 ; i<=n; i++){ ; j<=m; j++){ //注意这里的j是从0开始而非a[i] if(j>=a[i]){ dp[i][j]=max(dp[i-][j-a[i]]+b[i], dp[i-][j]); }els…
In Action Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5472    Accepted Submission(s): 1843 Problem Description Since 1945, when the first nuclear bomb was exploded by the Manhattan Project t…
题意:给你若干个集合,每个集合内的物品要么选任意一个,要么所有都选,求最后在背包能容纳的范围下最大的价值. 分析:对于每个并查集,从上到下滚动维护即可,其实就是一个01背包= =. 代码如下: #include <stdio.h> #include <algorithm> #include <string.h> #include <vector> using namespace std; + ; int w[N],b[N]; int n,m,W; int r…
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…