Collecting one's own plants for use as herbal medicines is perhaps one of the most self-empowering things a person can do, as it implies that they have taken the time and effort to learn about the uses and virtues of the plant and how it might benefi…

题目链接:http://acm.split.hdu.edu.cn/showproblem.php?pid=5887 题解:这题一看像是背包但是显然背包容量太大了所以可以考虑用dfs+剪枝,贪心得到的不一定是正确答案.当然这题还可以用背包来写,其实这就用到了dp的一些优化就是存状态,递推过程中有些状态是多余的没必要计算这样就可以大大减少空间的利用和时间的浪费 第一份是dfs+剪枝的写法第二份是背包+map存状态的写法. #include <iostream> #include <cstri…

http://acm.hdu.edu.cn/showproblem.php?pid=5887 题意: 容量很大的01背包. 思路: 因为这道题目背包容量比较大,所以用dp是行不通的.所以得用搜索来做,但是需要一些剪枝,先按体积排序,优先考虑体积大的物品,这样剪枝会剪得多一些(当然按照性价比排序也是可以的). #include<iostream> #include<algorithm> #include<cstring> #include<cstdio> #i…

背包,$map$,优化. 和普通背包一样,$map$加一个$erase$优化一下就可以跑的很快了. #pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<vector> #include<map> #…

Herbs Gathering 10.76% 1000ms 32768K   Collecting one's own plants for use as herbal medicines is perhaps one of the most self-empowering things a person can do, as it implies that they have taken the time and effort to learn about the uses and virtu…

http://acm.hdu.edu.cn/showproblem.php?pid=5808 用bitset<120>dp,表示dp[0] = true,表示0出现过,dp[100] = true表示100这个数字出现过. 对于每一个新的数字,val,有转移方程, dp = dp | (dp << val) 比如现在0000101,然后枚举了2进来,0000101 | 0010100 那个区间是暴力扫过去的,是水过去的,关键学了下bitset优化的背包. #include <…

Herbs Gathering Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 687    Accepted Submission(s): 145 Problem Description Collecting one's own plants for use as herbal medicines is perhaps one of t…

FATE Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 11908    Accepted Submission(s): 5645 Problem Description 最近xhd正在玩一款叫做FATE的游戏,为了得到极品装备,xhd在不停的杀怪做任务.久而久之xhd开始对杀怪产生的厌恶感,但又不得不通过杀怪来升完这最后一级.现在的…

HDU 1712 ACboy needs your help(包背包) pid=1712">http://acm.hdu.edu.cn/showproblem.php? pid=1712 题意: 小杰有m天的时间去上n门不同的课. 对于第i门课来说, 假设小杰花j天的时间在该课上, 那么小杰能够获得val[i][j]的价值. 如今给出矩阵val[n][m], 要你求出小杰能获得的最大价值和? 分析: 咋一看, n门课, m天的时间, 要我们求最大价值. 那么明显是从n门课中选出合适的几门,…

Dividing Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 53029   Accepted: 13506 Description Marsha and Bill own a collection of marbles. They want to split the collection among themselves so that both receive an equal share of the marbl…