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…

引用:http://szy961124.blog.163.com/blog/static/132346674201092775320970/ 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…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2639 题意:给出一行价值,一行体积,让你在v体积的范围内找出第k大的值 分析:dp[i][j][k]表示前i个物品容积为j时的第k优解.那么对于每种状态dp[i][j]都需要维护好前k优解. 每次根据前k优解进行每种取或不取第i件物品,用数组a记录取第i件物品,数组b记录不取,这样 数组a,b了所有能组成j的x种解,最后在x里取前k优解记录下来就好.剩下的肯定不是前k优解了... #include…

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…

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…

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…

Bone Collector II Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7165    Accepted Submission(s): 3802 Problem Description The title of this problem is familiar,isn't it?yeah,if you had took par…

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/…