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;…

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…

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…

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…

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…

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…

Bone Collector II Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3355    Accepted Submission(s): 1726 Problem Description The title of this problem is familiar,isn't it?yeah,if you had took par…

The title of this problem is familiar,isn't it?yeah,if you had took part in the "Rookie Cup" competition,you must have seem this title.If you haven't seen it before,it doesn't matter,I will give you a link: Here is the link: http://acm.hdu.edu.c…

Bone Collector II Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1334    Accepted Submission(s): 666 Problem Description The title of this problem is familiar,isn't it?yeah,if you had took par…

Bone Collector II Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4229    Accepted Submission(s): 2205 Problem Description The title of this problem is familiar,isn't it?yeah,if you had took par…

这题和典型的01背包求最优解不同,是要求第k优解,所以,最直观的想法就是在01背包的基础上再增加一维表示第k大时的价值.具体思路见下面的参考链接,说的很详细 参考连接:http://laiba2004.blog.163.com/blog/static/8835120220138611342496/http://hi.baidu.com/chenyun00/item/1c6c44318acc8bfaa88428c7 #include <iostream> #include <cstdio&…

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…

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…

题目链接 Problem Description The title of this problem is familiar,isn't it?yeah,if you had took part in the "Rookie Cup" competition,you must have seem this title.If you haven't seen it before,it doesn't matter,I will give you a link: Here is the l…

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&…

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…

题目大意 一个人收藏骨头,有 n 个骨头,每个骨头有体积和价值,问能够装在容量为 V 的背包中,能获得的第 k 大(去重后)价值是多少. 样例 样例输入 1 5 10 2 1 2 3 4 5 5 4 3 2 1 样例输出 1 12 样例输入 2 5 10 12 1 2 3 4 5 5 4 3 2 1 样例输出 2 2 分析 跑暴力显然不优秀,每种物品可选可不选,最多 \(2^n\) 种不同的方案,也就对应这么多价值,显然如果都存下再排序输出结果,简单了事,但是显然时间和空间都承受不住. 当然在跑…

Bone Collector II Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4178    Accepted Submission(s): 2174 Problem Description The title of this problem is familiar,isn't it?yeah,if you had took pa…

Bone Collector II Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5463    Accepted Submission(s): 2880 Problem Description The title of this problem is familiar,isn't it?yeah,if you had took par…

题目链接:pid=2602">HDU 2602 Bone Collector Bone Collector Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 28903    Accepted Submission(s): 11789 Problem Description Many years ago , in Teddy's h…

解题思路:对于01背包的状态转移方程式f[v]=max(f[v],f[v-c[i]+w[i]]);其实01背包记录了每一个装法的背包值,但是在01背包中我们通常求的是最优解, 即为取的是f[v],f[v-c[i]]+w[i]中的最大值,但是现在要求第k大的值,我们就分别用两个数组保留f[v]的前k个值,f[v-c[i]]+w[i]的前k个值,再将这两个数组合并,取第k名. 即f的数组会增加一维. http://blog.csdn.net/lulipeng_cpp/article/details/…

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…

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…

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…

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…

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[…

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…

题意:这题就是一个纯粹的裸01背包 分析:WA了好几次.01背包实现的一些细节没搞懂 1.为什么dp[i][j]赋初值为0而不是value[i].由于第i个石头可能不放! 2.在进行状态转移之前要dp[i][j]=dp[i-1][j],不然肯定会WA啊.想想就明确了 3.终于结果是dp[n][v],不是每次求mx,由于状态转移就是这么推的啊 代码: #include<iostream> #include<cstring> #include<cmath> using na…

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…