Bone Collector（复习01背包）

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HDOJ(HDU).2602 Bone Collector (DP 01背包）

HDOJ(HDU).2602 Bone Collector (DP 01背包) 题意分析 01背包的裸题 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #define nmax 1005 using namespace std; int v[nmax],w[nmax],dp[nmax]; int main() { //freopen("in…

hdu 2602 Bone Collector（01背包）模板

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2602 Bone Collector Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 54132 Accepted Submission(s): 22670 Problem Description Many years ago , in…

HDU2602 Bone Collector 【01背包】

Bone Collector Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 28365 Accepted Submission(s): 11562 Problem Description Many years ago , in Teddy's hometown there was a man who was called "Bo…

HDU 2639 Bone Collector II(01背包变形【第K大最优解】)

Bone Collector II Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4739 Accepted Submission(s): 2470 Problem Description The title of this problem is familiar,isn't it?yeah,if you had took pa…

hdu2602 Bone Collector（01背包） 2016-05-24 15:37 57人阅读评论(0) 收藏

Bone Collector Problem Description Many years ago , in Teddy's hometown there was a man who was called "Bone Collector". This man like to collect varies of bones , such as dog's , cow's , also he went to the grave - The bone collector had a big…

hdu 2602 - Bone Collector（01背包）解题报告

Bone Collector Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 39532 Accepted Submission(s): 16385 Problem Description Many years ago , in Teddy's hometown there was a man who was called "Bo…

题解报告：hdu 2602 Bone Collector（01背包）

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2602 Problem Description Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave…

HDU2602 Bone Collector（01背包）

HDU2602 Bone Collector 01背包模板题 #include<stdio.h> #include<math.h> #include<string.h> #include<stdlib.h> #define maxn 1005 int w[maxn], v[maxn], f[maxn]; int N, V, T; int _max(int a, int b){ if(a > b) return a; else return b; } i…

hdu–2369 Bone Collector II(01背包变形题)

题意:求解01背包价值的第K优解. 分析: 基本思想是将每个状态都表示成有序队列,将状态转移方程中的max/min转化成有序队列的合并. 首先看01背包求最优解的状态转移方程:\[dp\left[ j \right] = \max \left\{ {dp\left[ j \right],dp\left[ {j - a\left[ i \right].w} \right] + a\left[ i \right].v} \right\}\] 如果要求第K优解,那么状态 dp[j] 就应该是一个大小为…

Bone Collector（01背包）

题目链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87125#problem/N 题目: Description Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also h…

hdu 2602 Bone Collector（01背包）

题意:给出包裹的大小v,然后给出n块骨头的价值value和体积volume,求出一路下来包裹可以携带骨头最大价值思路:01背包 1.二维数组(不常用 #include<iostream> #include<stdio.h> #include<math.h> using namespace std; ][]; int main() { int i,j; int t,n,_v;//测试用例个数,物品种类,背包大小 ],v[];//花费,价值 scanf("%d&…

hdoj 2602 Bone Collector 【01背包】

意甲冠军:给出的数量和袋骨骼的数,然后给每块骨骼的价格值和音量.寻求袋最多可容纳骨骼价格值难度;这个问题是最基本的01背包称号,不知道的话,推荐看<背包9说话> AC by SWS 主题链接 http://acm.hdu.edu.cn/showproblem.php?pid=2602 代码: #include<stdio.h> #include<string.h> typedef struct{ int w, v; }str; str s[1005]; int dp[…

Hdu2602 Bone Collector （01背包）

Problem Description Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave …The bone collector had a big bag with a volume of…

HDU 2639 Bone Collector II (01背包，第k解)

题意: 数据是常规的01背包,但是求的不是最大容量限制下的最佳解,而是第k佳解. 思路: 有两种解法: 1)网上普遍用的O(V*K*N). 2)先用常规01背包的方法求出背包容量限制下能装的最大价值m,再以m为背包容量再进行一次01背包,dp[j]表示当物品的组合价值为j时,它们的体积之和的最小量.那么就求出了所有可能的价值,从1-m都有,但是其中一些是求不出来的,也就是骨头的价值不能组合成这个数字,那么就得过滤掉. #include <iostream> #include <cstdi…

HDU - 2602 Bone Collector（01背包讲解）

题意:01背包:有N件物品和一个容量为V的背包.每种物品均只有一件.第i件物品的费用是volume[i],价值是value[i],求解将哪些物品装入背包可使价值总和最大. 分析: 1.构造二维数组: dp[i][j]---前i件物品放入一个容量为j的背包可以获得的最大价值. dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - volume[i]] + value[i]);---(a) (1)dp[i - 1][j]---不放第i件物品,因此前i件物品放入一个容量…

HDU 2639 Bone Collector II(01背包变型)

此题就是在01背包问题的基础上求所能获得的第K大的价值. 详细做法是加一维去推当前背包容量第0到K个价值,而这些价值则是由dp[j-w[ i ] ][0到k]和dp[ j ][0到k]得到的,事实上就是2个数组合并之后排序,可是实际做法最好不要怎么做.由于你不知道总共同拥有多少种.而我们最多仅仅须要前K个大的即可了(由于可能2个数组加起来的组合数达不到K个),假设所有加起来数组开多大不清楚,所以能够选用归并排序中把左右2个有序数组合并成一个有序数组的方法来做.就是用2个变量去标记2个有序数组的头…

杭电2602 Bone Collector 【01背包】

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2602 解题思路:给出一个容量为V的包,以及n个物品,每一个物品的耗费的费用记作c[i](即该物品的体积),每一个物品的价值记作w[i], 我们用 f[v]来表示一个容量为v的包的总价值,这样最后我们只需要输出f[V]就能得出结果则对于第i个物品,它可以放入背包,此时背包的容量变为v-c[i],背包的总价值变为f[v-c[i]]+w[i], 它也可以不放入背包,此时背包的容量还是v,背包的总价值不变…

[HDOJ2602]Bone Collector（01背包）

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2602 裸的... #include <algorithm> #include <iostream> #include <iomanip> #include <cstring> #include <climits> #include <complex> #include <fstream> #include <casser…

HDU - 2639 Bone Collector II (01背包第k大解)

分析 \(dp[i][j][k]\)为枚举到前i个物品,容量为j的第k大解.则每一次状态转移都要对所有解进行排序选取前第k大的解.用两个数组\(vz1[],vz2[]\)分别记录所有的选择情况,并选择其中前k大的更新当前的dp[i][k].因为dp[i]满足递增的特点,所以可以对两个数组顺序比较选择. #include<bits/stdc++.h> using namespace std; const int maxn = 1e3+5; const int INF = 0x3f3f3f3f;…