http://acm.hdu.edu.cn/showproblem.php?pid=2639 在背包的基础上维护一个size<=K的最大值集合,为什么维护K个就好了呢,因为如果当前状态有多余K个最优解,前K个就足够转移到下一状态并占满前K了,所以K个之后的都没必要维护. #include <iostream> #include <cstdio> #include <vector> #include <string> #include <map&g…

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…

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…

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=2639 题意:求01背包的第k优解 dp(i, j)表示容量为j时的i优解 对于第二维的操作和01背包几乎一样,只是我们只需要关注每一次取的前k大个即可. #include <algorithm> #include <iostream> #include <iomanip> #include <cstring> #include <climits> #…

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…

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…

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;…

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 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 60469    Accepted Submission(s): 25209 Problem Description Many years ago , in Teddy’s hometown there was a man who was called “Bo…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2602 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 collect…

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/…

这种01背包的裸题,本来是不想写解题报告的.但是鉴于还没写过背包的解题报告.于是来一发. 这个真的是裸的01背包. 代码: #include <iostream> #include <cstdio> using namespace std; #define N 1007 int c[N],w[N],dp[N]; int main() { int t,i,n,V,v; scanf("%d",&t); while(t--) { scanf("%d%…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2602 水题啊水题 还给我WA了好多次 因为我在j<w[i]的时候状态没有下传.. #include <cstdio> #include <algorithm> #include <cstring> using namespace std; typedef long long LL; typedef pair<int,int> PII; #define PB…

题意:给定一个体积,和一些物品的价值和体积,问你最大的价值. 析:最基础的01背包,dp[i] 表示体积 i 时最大价值. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream>…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2639 题意: 01背包第k优解, 背包九讲原题.“ 对于求次优解.第K优解类的问题,如果相应的最优解问题能写出状态转移方程.用动态规划解决,那么求次优解往往可以相同的复杂度解决,第K优解则比求最优解的复杂度上多一个系数K. 其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.这里仍然以01背包为例讲解一下. 首先看01背包求最优解的状态转移方程:f[i][v…

Bone Collector II Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3471    Accepted Submission(s): 1792 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): 4824    Accepted Submission(s): 2514 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…

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…

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…

/* 01背包第k优解问题 f[i][j][k] 前i个物品体积为j的第k优解 对于每次的ij状态 记下之前的两种状态 i-1 j-w[i] (选i) i-1 j (不选i) 分别k个 然后归并排序并且去重生成ij状态的前k优解 */ #include<iostream> #include<cstdio> #include<cstring> #define maxn 1010 using namespace std; ],x[maxn],y[maxn],a,b,z; i…

Bone Collector II Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3042 Accepted Submission(s): 1578 Problem Description The title of this problem is familiar,isn't it?yeah,if you had took part in…

http://acm.hdu.edu.cn/showproblem.php?pid=2639       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 does…