Bone Collector 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…

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…

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…

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…

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 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…

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求01背包的第k大解.合并两个有序序列 选取物品i,或不选.最终的结果,是我们能在O(1)的时间内,判定对于体积j,是否应当选取第i件物品. 我们在这里作出了最优的选择.那被我们抛弃的选择呢?他很可能是次优解,第三优解,无论怎样,他都对我们本题求前K优解,起到了重要的作用! #include<stdio.h> #include<string.h> #include<algor…

传送门 题目大意:01背包裸题. 复习01背包: 题目 有N件物品和一个容量为V的背包.第i件物品的费用是c[i],价值是w[i].求解将哪些物品装入背包可使这些物品的费用总和不超过背包容量,且价值总和最大. 题解: #include<iostream> #include<cstdio> #include<cstring> using namespace std; ],w[],v[]; int n,m,t; int main() { scanf("%d&quo…

题目http://acm.hdu.edu.cn/showproblem.php?pid=2639 分析:这是求第K大的01背包问题,很经典.dp[j][k]为背包里面装j容量时候的第K大的价值. 从普通01背包中可以知道最大价值dp[j]是由dp[j]和dp[j-c[i]]+w[i]决定的.那么可以知道 dp[j][k]也是有dp[j][k]和dp[j-c[i]][k]来决定的. 求次优解.第K优解: 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…